###贪心证明
假设相邻的两个人左右手分别是(a, b), (A, B)。设a * b <= A * B，i之前所有人的左手乘积为S。
则，ans1 = max{S / b, S * a / B}
若交换
则，ans2 = max{S / B, S * A / b}
因为，a * b <= A * B
所以，S * a / B <= S * A / b
又因为，S / B <= S * a / B
所以，ans2 = S * A / b
ans1 = max{S / B[i], S * a / B}
所以，ans1 <= ans2
证毕
###Code
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
using namespace std;
#define pii pair<int, int>
#define fi first
#define se second
#define piii pair<int, pii >
const int maxn = 1000 + 2, Len = 4000 + 1, base = 10000;
struct bint
{
int len;
int v[Len];
bint(int r = 0) {for (len = 0; r > 0; r /= base) v[len++] = r % base;}
bint &operator = (const bint &a)
{
memcpy(this, &a, sizeof(int) * (a.len + 1));
return *this;
}
bool operator < (const bint &a)
{
if (len != a.len) return len < a.len;
for (int i = len - 1; i >= 0; i--)
if (v[i] != a.v[i]) return v[i] < a.v[i];
return false;
}
};
ostream &operator << (ostream &out, const bint &a)
{
printf("%d", a.len == 0 ? 0 : a.v[a.len - 1]);
for (int i = a.len - 2; i >= 0; i--) printf("%04d", a.v[i]);
return out;
}
bint operator * (const bint &a, const int &b)
{
bint ans = 0;
if (!a.len || !b) return ans;
int carry = 0, i;
for (i = 0; i < a.len || carry; i++)
{
if (i < a.len) carry += a.v[i] * b;
ans.v[i] = carry % base;
carry /= base;
}
ans.len = i;
return ans;
}
bint operator / (const bint &a, const int &b)
{
bint ans = a;
int carry = 0;
for (int i = ans.len - 1; i >= 0; i--)
{
ans.v[i] += carry * base;
carry = ans.v[i] % b;
ans.v[i] /= b;
}
while (!ans.v[ans.len - 1] && ans.len) ans.len--;
return ans;
}
inline void gmax(bint &a, const bint &b) {if (a < b) a = b;}
int n, a, b;
piii d[maxn];
int main()
{
scanf("%d%d%d", &n, &a, &b);
for (int i = 1; i <= n; i++)
scanf("%d%d", &d[i].se.fi, &d[i].se.se), d[i].fi = d[i].se.fi * d[i].se.se;
sort(d + 1, d + n + 1);
bint to = a, ans = 0;
for (int i = 1; i <= n; i++)
{
gmax(ans, to / d[i].se.se);
to = to * d[i].se.fi;
}
cout << ans << endl;
return 0;
}

简单高精度
#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int MaxN=1001;
struct NodeTp {
int L;
int R;
int W;
} p[MaxN];
int n,KL,KR;
int a[MaxN],b[MaxN],ans[MaxN];
//比较大小
int ncmp() {
int i;
if (b[0]!=ans[0])
return b[0]-ans[0];
else {
for (i=b[0]; i>=1; i--)
if (b[i]!=ans[i])
return b[i]-ans[i];
return 0;
}
}
//输出结果
void out() {
int i;
printf("%d",ans[ans[0]]);
for (i=ans[0]-1; i>=1; i--)
printf("%4.4d",ans[i]);
}
//高精度*低精度
void mult(int x) {
int i,k=0,tmp;
for (i=1; i<=a[0]||k; i++) {
tmp=a[i]*x+k;
k=tmp/10000;
a[i]=tmp%10000;
}
a[0]=i-1;
}
//高精度/低精度
void div(int x) {
int i,k=0,tmp;
for (i=a[0]; i>=1; i--) {
tmp=a[i]+k*10000;
b[i]=tmp/x;
k=tmp%x;
}
for (i=a[0]; i>1&&!b[i]; i--);
b[0]=i;
}
int cmp(NodeTp a,NodeTp b) {
return a.L*a.R<b.L*b.R;
}
void init() {
int i;
scanf("%d%d%d",&n,&KL,&KR);
for (i=1; i<=n; i++) {
scanf("%d%d",&p[i].L,&p[i].R);
p[i].W=p[i].L*p[i].R;
}
sort(p+1,p+n+1,cmp);
}
void work() {
int i;
a[0]=1,a[1]=KL;
ans[0]=1,ans[1]=0;
for (i=1; i<=n; i++) {
div(p[i].R);
if (ncmp()>0)
memcpy(ans,b,sizeof(b));
mult(p[i].L);
}
}
int main() {
init();
work();
out();
return 0;
}

Python裸题
python
#!/bin/python3
n = int(input())
ka = int(input().split()[0])
z = []
for i in range(0, n):
z.append(list(map(int, input().split())))
z.sort(key = lambda d: d[0] * d[1])
for i in z[0 : n - 1]:
ka = ka * i[0]
print(ka // z[n - 1][1])


#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#include <vector>
#include <deque>
#include <set>
#include <limits>
#include <string>
#include <sstream>
using namespace std;

const int oo_min=0xcfcfcfcf,oo_max=0x3f3f3f3f;
const int max_size=4000;

struct p_1
{
    int a,b,c;
}p[1000+1];

int n,ans_size;
int ans[max_size];
int sum[max_size];
int temp[max_size];

int cmp_p_1_1(p_1 x,p_1 y)
{
    return x.c<y.c;
}

int get_size(int *a)
{
    for (int i=max_size;i>=1;i--)
        if (a[i-1]!=0)
            return i;
    return 0;
}

int cmp_a_1(int *a,int *b)
{
    int a_size=get_size(a),b_size=get_size(b);
    if (a_size<b_size)
        return 1;
    else if (a_size>b_size)
        return 0;
    else
    {
        for (int i=a_size-1;i>=0;i--)
            if (a[i]<b[i])
                return 1;
            else if (a[i]>b[i])
                return 0;
        return 0;
    }
}

void calc_1(int *a,int b,int *c)
{
    memset(c,0,sizeof(c));
    int a_size=get_size(a);
    for (int i=0;i<a_size;i++)
        c[i]=a[i]*b;
    for (int i=0;i<a_size;i++)
    {
        if (c[i]>=10&&i==a_size-1)
            a_size++;
        c[i+1]+=c[i]/10;
        c[i]%=10;
    }
}

void calc_2(int *a,int b,int *c)
{
    memset(c,0,sizeof(c));
    int a_size=get_size(a);
    for (int i=a_size-1,temp=0;i>=0;i--)
    {
        temp=(temp*10)+a[i];
        c[i]=temp/b;
        temp%=b;
    }
}

int main()
{
    if (~scanf("%d",&n))
    {
        for (int i=0;i<=n;i++)
        {
            scanf("%d%d",&p[i].a,&p[i].b);
            p[i].c=p[i].a*p[i].b;
        }
        sort(&p[1],&p[n+1],cmp_p_1_1);
        memset(ans,0,sizeof(ans));
        memset(sum,0,sizeof(sum));
        sum[0]=1;
        for (int i=0;i<=n;i++)
        {
            if (i>0)
            {
                memset(temp,0,sizeof(temp));
                calc_2(sum,p[i].b,temp);
                if (cmp_a_1(ans,temp))
                    memcpy(ans,temp,sizeof(ans));
            }
            memset(temp,0,sizeof(temp));
            calc_1(sum,p[i].a,temp);
            memcpy(sum,temp,sizeof(sum));
        }
        ans_size=get_size(ans);
        for (int i=ans_size-1;i>=0;i--)
            printf("%d",ans[i]);
        printf("\n");
    }
}

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

struct peo{
int a,b,ab;
}r[1001];

int c[1001],q[1001],n,i,j,len,len1,kinga,kingb,ans[1001],lena;

bool cmp(peo x,peo y)
{
return x.ab<y.ab;
}

void bj()
{
int i;
if (lena>len1) return;
else
if (lena==len1)
{
for (i=lena; i>=1; i--)
if (q[i]<ans[i]) return;
else if (q[i]>ans[i]) break;
if (i==0) return;
}
for (i=len1; i>=1; i--)
ans[i]=q[i];
lena=len1;
}

void chu(int k)
{
for (j=len1; j>=2; j--)
q[j-1]+=q[j]%r[i].b*10000,q[j]/=r[i].b;
q[1]/=r[i].b;
while (q[len1]==0)
len1--;
}

void cheng(int k)
{
for (j=1; j<=len; j++)
c[j]*=r[i].a;
for (j=1; j<=len; j++)
if (c[j]>10000)
c[j+1]+=c[j]/10000,c[j]%=10000;
while (c[len+1]>0)
len++,c[len+1]+=c[len]/10000,c[len]%10000;
}

int main()
{
scanf("%d",&n);
scanf("%d %d",&kinga,&kingb);
for (i=1; i<=n; i++)
scanf("%d %d",&r[i].a,&r[i].b),r[i].ab=r[i].a*r[i].b;
sort(r+1,r+n+1,cmp);
c[1]=kinga; len=1;
for (i=1; i<=n; i++)
{
memcpy(q,c,sizeof(c)),len1=len;
chu(r[i].b);
bj();
cheng(r[i].a);
}
printf("%d",ans[lena]);
for (j=lena-1; j>=1; j--)
{
if (ans[j]<1000) printf("0");
if (ans[j]<100) printf("0");
if (ans[j]<10) printf("0");
printf("%d",ans[j]);
}
return 0;
}

贪心算法足矣．
注意不考虑大数只有60分哦~

import java.math.BigInteger;
import java.util.Arrays;
import java.util.Scanner;

public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[] king = { sc.nextInt(), sc.nextInt() };
int[][] dc = new int[n][2];
for (int i=0; i<n; i++) {
dc[i][0] = sc.nextInt();
dc[i][1] = sc.nextInt();
}

for (int i=0; i<n-1; i++) {
for (int j=i + 1; j<n; j++) {
int a = dc[i][0];
int b = dc[i][1];

if( Math.max(dc[j][1], a*b) > Math.max(b, dc[j][0]*dc[j][1]) ) {
dc[i][0] = dc[j][0];
dc[i][1] = dc[j][1];
dc[j][0] = a;
dc[j][1] = b;
}

}
}

BigInteger money = new BigInteger("0");
BigInteger s = new BigInteger(king[0] + "");

for (int i=0; i<n; i++) {
BigInteger dci1 = new BigInteger(dc[i][1] + "");
BigInteger sDiv = s.divide(dci1);

if (sDiv.compareTo( money ) == 1) {
money = sDiv;
}

BigInteger dci0 = new BigInteger(dc[i][0] + "");
s = s.multiply(dci0);
}

System.out.println(money.toString());

}
}

高中时代做的题拿java水一发

import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Scanner;

class People implements Comparable<People> {
    BigInteger a, b;

    public BigInteger getA() {
        return a;
    }

    public void setA(BigInteger a) {
        this.a = a;
    }

    public BigInteger getB() {
        return b;
    }

    public void setB(BigInteger b) {
        this.b = b;
    }

    //定义compareTo方法来使用Collections.sort()
    @Override
    public int compareTo(People rhs) {
        return a.multiply(b).subtract(rhs.a.multiply(rhs.b)).compareTo(BigInteger.valueOf(0));
    }
}

public class Main {
    private static BigInteger max(BigInteger a, BigInteger b) {
        if (a.compareTo(b) < 0) return b;
        else return a;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n;
        People king = new People();
        People buf;
        ArrayList<People> p = new ArrayList<>();
        while (sc.hasNext()) {
            n = sc.nextInt();
            king.setA(sc.nextBigInteger());
            king.setB(sc.nextBigInteger());
            for (int i = 0; i < n; i++) {
                buf = new People(); //ArrayList.add()方法传的是引用，所以要保证每次初始化一个新的对象
                buf.setA(sc.nextBigInteger());
                buf.setB(sc.nextBigInteger());
                p.add(buf);
            }
            Collections.sort(p);
            BigInteger ans = BigInteger.valueOf(0);
            BigInteger sum = king.getA();
            for (int i = 0; i < n; i++) {
                ans = max(ans, sum.divide(p.get(i).getB()));
                sum = sum.multiply(p.get(i).getA());
            }
            System.out.println(ans);
        }
    }
}

//60分做法
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int MAXN = 1000 + 5;
int n,a[MAXN],b[MAXN];
struct c{
    int a,b;
    bool operator < (const c & rhs) const {
        return (long long)a*b < (long long)rhs.a*rhs.b;
    }
}c[MAXN];
int main(){
    int n;
    scanf("%d",&n);
    for(int i=0;i<=n;i++){
        scanf("%d%d",&c[i].a,&c[i].b);
    }
    sort(c+1,c+1+n);
    int ans=0,sum=c[0].a;
    for(int i=1;i<=n;i++){
        ans=max(ans,sum/c[i].b);
        sum*=c[i].a;
    }
    printf("%d\n",ans);
    return 0;
    
}

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int MaxN=1001;
struct NodeTp {
int L;
int R;
int W;
} p[MaxN];
int n,KL,KR;
int a[MaxN],b[MaxN],ans[MaxN];
//比较大小
int ncmp() {
int i;
if (b[0]!=ans[0])
return b[0]-ans[0];
else {
for (i=b[0]; i>=1; i--)
if (b[i]!=ans[i])
return b[i]-ans[i];
return 0;
}
}
//输出结果
void out() {
int i;
printf("%d",ans[ans[0]]);
for (i=ans[0]-1; i>=1; i--)
printf("%4.4d",ans[i]);
}
//高精度*低精度
void mult(int x) {
int i,k=0,tmp;
for (i=1; i<=a[0]||k; i++) {
tmp=a[i]*x+k;
k=tmp/10000;
a[i]=tmp%10000;
}
a[0]=i-1;
}
//高精度/低精度
void div(int x) {
int i,k=0,tmp;
for (i=a[0]; i>=1; i--) {
tmp=a[i]+k*10000;
b[i]=tmp/x;
k=tmp%x;
}
for (i=a[0]; i>1&&!b[i]; i--);
b[0]=i;
}
int cmp(NodeTp a,NodeTp b) {
return a.L*a.R<b.L*b.R;
}
void init() {
int i;
scanf("%d%d%d",&n,&KL,&KR);
for (i=1; i<=n; i++) {
scanf("%d%d",&p[i].L,&p[i].R);
p[i].W=p[i].L*p[i].R;
}
sort(p+1,p+n+1,cmp);
}
void work() {
int i;
a[0]=1,a[1]=KL;
ans[0]=1,ans[1]=0;
for (i=1; i<=n; i++) {
div(p[i].R);
if (ncmp()>0)
memcpy(ans,b,sizeof(b));
mult(p[i].L);
}
}
int main() {
init();
work();
out();
return 0;
}

#include <cstdio>
#include <iostream>
using namespace std;

const int MaxN=1001;

struct NodeTp
{int L;//左手上的数
int R;//右手上的数
int W;//两者乘积
}p[MaxN];

int n,KL,KR;
int a[MaxN],b[MaxN],ans[MaxN];

int ncmp()//比较两个高精度数的大小
{int i;
if (b[0]!=ans[0])
return b[0]-ans[0];
else
{for (i=b[0];i>=1;i--)
if (b[i]!=ans[i]) return b[i]-ans[i];
return 0;
}
}

void out()//输出:数组的每一位存4位数(即万进制)
{int i;
printf("%d",ans[ans[0]]);
for (i=ans[0]-1;i>=1;i--)
printf("%4.4d",ans[i]);
}

void mult(int x)//高精度乘单精度数x
{int i,k=0,tmp;
for (i=1;i<=a[0]||k;i++)
{tmp=a[i]*x+k;
k=tmp/10000;
a[i]=tmp%10000;

}
a[0]=i-1;//a[0]存位数
}

void div(int x)//高精度除以单精度数x
{int i,k=0,tmp;
for (i=a[0];i>=1;i--)
{tmp=a[i]+k*10000;
b[i]=tmp/x;
k=tmp%x;
}
for (i=a[0];i>1&&!b[i];i--);
b[0]=i;//b[0]存位数
}

int cmp(const void* a,const void* b)
{NodeTp x=*(NodeTp )a;
NodeTp y=(NodeTp *)b;
return x.W > y.W ? 1:-1;
}

void init()
{int i;
scanf("%d%d%d",&n,&KL,&KR);
for (i=1;i<=n;i++)
{scanf("%d%d",&p[i].L,&p[i].R);
p[i].W=p[i].L*p[i].R;
}
qsort(p+1,n,sizeof(p[1]),cmp);//按左手数*右手数从小到大排序
}

void work()
{int i;
a[0]=1,a[1]=KL;
ans[0]=1,ans[1]=0;
for (i=1;i<=n;i++)
{div(p[i].R);//b[]记录第i个大臣获得的金币数
if (ncmp()>0) memcpy(ans,b,sizeof(b));
mult(p[i].L);////a[]记录从国王到第i个大臣左手上的数的乘积

}
}

int main()
{freopen("game.in","r",stdin);
freopen("game.out","w",stdout);
init();

work();
out();
return 0;
}

高精度数组记得要定大点，至少4001，为了凑个整数我定了5000……

#include <iostream>
#include <cstring>
using namespace std;
long l[5001],ans[5001],Max[5001];
long len;
void cf(long x)
{
    long i,g;
    g=0;
    for (i=1;i<=len;i++){
        l[i]=l[i]*x+g;
        g=l[i]/10;
        l[i]=l[i]%10;
    }
    while (g!=0){
        l[i]=l[i]+g;
        g=l[i]/10;
        l[i]=l[i]%10;
        i++;
    }
    len=i-1;
}
void c(long x)
{
    long i,g;
    memset(ans,0,sizeof(ans));
    g=0;
    for (i=len;i>=1;i--){
        ans[i]=(g*10+l[i])/x;
        g=(g*10+l[i])%x;
    }
}
void bj()
{
    long i;
    for (i=len;i>=1;i--)
        if (Max[i]<ans[i]){
            memcpy(Max,ans,sizeof(ans));
            break;
        } else if (Max[i]>ans[i])break;
}
main()
{
    long a[1001],b[1001],i,j,k,n;
    cin>>n;
    for (i=0;i<=n;i++)
        cin>>a[i]>>b[i];
    for (i=1;i<n;i++)
        for (j=i+1;j<=n;j++)
            if (a[i]*b[i]>a[j]*b[j]){
                k=a[i];
                a[i]=a[j];
                a[j]=k;
                k=b[i];
                b[i]=b[j];
                b[j]=k;
            }
    memset(l,0,sizeof(l));
    len=0;
    while (a[0]!=0){
        len++;
        l[len]=a[0]%10;
        a[0]=a[0]/10;
    }
    for (i=1;i<=n;i++){
        c(b[i]);
        bj();
        cf(a[i]);
    }
    len=5000;
    while (Max[len]==0) len--;
    for (i=len;i>=1;i--) cout<<Max[i];
}

Vijos 题解：http://hi.baidu.com/umule/item/2c997f8ed9600fdae596e017
有疑问请留言 共同进步

可算AC了
卡死我了
vijos数据太弱了
没有比较数组都能过

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
using namespace std;
int n,t,z,nmp,tmp,k,ans;
int x[10000];
int y[10000];
int cheng[10000];
int lx;int ly;
struct your
{
int r,e;
}q[1011];
bool cmp(your m,your n)
{
if(m.r*m.e==n.r*n.e)
{
return m.e<n.e;
}
else
{
return m.r*m.e<n.r*n.e;
}
}
void init()
{
memset(x,0,sizeof x);
lx=0;tmp=0;nmp=0;ans=0;
return;
}
int main()
{
scanf("%d",&n);
for(int i=1;i<=n+1;i++)
{
scanf("%d%d",&q[i].r,&q[i].e);
}
sort(q+2,q+n+2,cmp);
t=1;
cheng[1]=1;
for(int w=2;w<=n+1;w++)
{
tmp=0;nmp=0;
for(int j=1;j<=t;j++)
{
tmp=cheng[j]*q[w-1].r+nmp;
cheng[j]=tmp%10;
nmp=tmp/10;
if(j==t&&nmp>0)
{
t++;
}
}
k=0;
for(int i=t;i>=1;i--)
{
ans=ans*10+cheng[i];
if(ans/q[w].e!=0){k=1;}
if(k==1)
{

lx++;
x[lx]=ans/q[w].e;
}
ans=ans%q[w].e;
}
int jk=0;
if(ly<lx)
{
for(int z=lx;z>=0;z--) y[z]=x[z];
ly=lx;
}
else
{
if(ly==lx)
{
for(int v=1;v<=lx;v++)
{
if(y[v]<x[v])
{
jk=1;
break;
}
}
if(jk==1)
{
for(int z=lx;z>=0;z--) y[z]=x[z];
jk=0;
}
}
}

init();
}
for(int i=1;i<=ly;i++)
{
printf("%d",y[i]);
}
}

type dd=array[0..5000] of longint;
var
a,b,p,d,py,ans,c:dd;
n,i,j,t:longint;

function time(a:dd; b:longint):dd;
var i,len:longint;
begin
len:=a[0]; fillchar(c,sizeof(c),0);
for i:=1 to len do
c[i]:=c[i]+a[i]*b;
for i:=1 to len do
begin
c[i+1]:=c[i+1]+c[i] div 10;
c[i]:=c[i] mod 10;
end;
while c[len+1]>0 do
begin
inc(len);
c[len+1]:=c[len] div 10;
c[len]:=c[len] mod 10;
end;
c[0]:=len;
time:=c;
end;

function divide(a:dd;b:longint):dd;
var i,len,d:longint;
begin
len:=a[0]; fillchar(c,sizeof(c),0);
d:=0;
for i:=len downto 1 do
begin
d:=d*10+a[i];
c[i]:=d div b;
d:=d mod b;
end;
while (len>=1) and (c[len]=0) do dec(len);
c[0]:=len;
divide:=c;
end;

function max(a,b:dd):boolean;
var i:longint;
begin
if a[0]>b[0] then exit(true);
if a[0]<b[0] then exit(false);
for i:=a[0] downto 1 do
if a[i]>b[i] then exit(true)
else if a[i]<b[i] then exit(false);
exit(false);
end;
begin
readln(n); py[0]:=1; py[1]:=1;
readln(a[0],b[0]); py:=time(py,a[0]);
for i:=1 to n do
begin
readln(a[i],b[i]);
p[i]:=a[i]*b[i];
end;
for i:=1 to n-1 do
for j:=i+1 to n do
if p[j]<p[i] then
begin
t:=a[i]; a[i]:=a[j]; a[j]:=t;
t:=b[i]; b[i]:=b[j]; b[j]:=t;
t:=p[i]; p[i]:=p[j]; p[j]:=t;
end;
ans[0]:=0;
for i:=1 to n do
begin
d:=divide(py,b[i]);
if max(d,ans) then ans:=d;
py:=time(py,a[i]);
end;
for i:=ans[0] downto 1 do
write(ans[i]);
end.

一个用vector实现的高精度
C++
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
struct Wint:vector<int>
{
Wint(const int &n=0)
{
push_back(n);
check();
}
Wint& check()
{
while(!empty()&&!back())pop_back();
if(empty())return *this;
for(int i=1; i!=size(); ++i)
{
(*this)[i]+=(*this)[i-1]/10;
(*this)[i-1]%=10;
}
while(back()>=10)
{
push_back(back()/10);
(*this)[size()-2]%=10;
}
return *this;
}
};
ostream& operator<<(ostream &os,const Wint &w)
{
for(int i=w.size()-1; i!=-1; --i)os<<w[i];
return w.empty()?os<<0:os;
}
bool operator<(const Wint &a,const Wint &b)
{
if(a.size()!=b.size())return a.size()<b.size();
for(int i=a.size()-1; i!=-1; --i)
if(a[i]!=b[i])return a[i]<b[i];
return 0;
}
Wint& operator*=(Wint &w,const int &n)
{
for(int i=0; i!=w.size(); ++i)w[i]*=n;
return w.check();
}
Wint operator/(Wint w,const int &n)
{
for(int i=w.size()-1,mod=0; i!=-1; --i)
{
w[i]+=mod*10;
mod=w[i]%n;
w[i]/=n;
}
return w.check();
}
struct People
{
int a,b;
};
bool operator<(const People &m,const People &n)
{
if(m.a*m.b!=n.a*n.b)return m.a*m.b<n.a*n.b;
if(m.a!=n.a)return m.a<n.a;
return 0;
}
int main()
{
vector<People> v(1);
int n;
Wint s=1,ans=0;
cin>>n>>v.back().a>>v.back().b;
while(n--)
{
v.push_back(v.front());
cin>>v.back().a>>v.back().b;
}
sort(v.begin()+1,v.end());
for(int i=1; i!=v.size(); ++i)
{
s*=v[i-1].a;
ans=max(ans,s/v[i].b);
}
cout<<ans;
}


不好意思，头文件没粘上来
#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<cstring>
using namespace std;
const int MaxN=1001;
struct NodeTp
{int L;
int R;
int W;
}p[MaxN];
int n,KL,KR;
int a[MaxN],b[MaxN],ans[MaxN];
int ncmp()
{int i;
if (b[0]!=ans[0])
return b[0]-ans[0];
else
{for (i=b[0];i>=1;i--)
if (b[i]!=ans[i]) return b[i]-ans[i];
return 0;
}
}

void out()
{int i;
printf("%d",ans[ans[0]]);
for (i=ans[0]-1;i>=1;i--)
printf("%4.4d",ans[i]);
}

void mult(int x)
{int i,k=0,tmp;
for (i=1;i<=a[0]||k;i++)
{tmp=a[i]*x+k;
k=tmp/10000;
a[i]=tmp%10000;

}
a[0]=i-1;
}

void div(int x)
{int i,k=0,tmp;
for (i=a[0];i>=1;i--)
{tmp=a[i]+k*10000;
b[i]=tmp/x;
k=tmp%x;
}
for (i=a[0];i>1&&!b[i];i--);
b[0]=i;
}

int cmp(const void* a,const void* b)
{NodeTp x=*(NodeTp )a;
NodeTp y=(NodeTp *)b;
return x.W > y.W ? 1:-1;
}
void init()
{int i;
scanf("%d%d%d",&n,&KL,&KR);
for (i=1;i<=n;i++)
{scanf("%d%d",&p[i].L,&p[i].R);
p[i].W=p[i].L*p[i].R;
}
qsort(p+1,n,sizeof(p[1]),cmp);
}
void work()
{int i;
a[0]=1,a[1]=KL;
ans[0]=1,ans[1]=0;
for (i=1;i<=n;i++)
{div(p[i].R);
if (ncmp()>0) memcpy(ans,b,sizeof(b));
mult(p[i].L);
}
}
int main()
{
init();

work();
out();
return 0;
}

高精度数组记得要定大点，至少4001，为了凑个整数我定了5000……
#include <iostream>
#include <cstring>
using namespace std;
long l[5001],ans[5001],Max[5001];
long len;
void cf(long x)
{
long i,g;
g=0;
for (i=1;i<=len;i++){
l[i]=l[i]*x+g;
g=l[i]/10;
l[i]=l[i]%10;
}
while (g!=0){
l[i]=l[i]+g;
g=l[i]/10;
l[i]=l[i]%10;
i++;
}
len=i-1;
}
void c(long x)
{
long i,g;
memset(ans,0,sizeof(ans));
g=0;
for (i=len;i>=1;i--){
ans[i]=(g*10+l[i])/x;
g=(g*10+l[i])%x;
}
}
void bj()
{
long i;
for (i=len;i>=1;i--)
if (Max[i]<ans[i]){
memcpy(Max,ans,sizeof(ans));
break;
} else if (Max[i]>ans[i])break;
}
main()
{
long a[1001],b[1001],i,j,k,n;
cin>>n;
for (i=0;i<=n;i++)
cin>>a[i]>>b[i];
for (i=1;i<n;i++)
for (j=i+1;j<=n;j++)
if (a[i]*b[i]>a[j]*b[j]){
k=a[i];
a[i]=a[j];
a[j]=k;
k=b[i];
b[i]=b[j];
b[j]=k;
}
memset(l,0,sizeof(l));
len=0;
while (a[0]!=0){
len++;
l[len]=a[0]%10;
a[0]=a[0]/10;
}
for (i=1;i<=n;i++){
c(b[i]);
bj();
cf(a[i]);
}
len=5000;
while (Max[len]==0) len--;
for (i=len;i>=1;i--) cout<<Max[i];
}

快排+高精乘+高精除
var
i,n,l:longint;
a,b,c:array[0..10000] of longint;
g,g2:array[0..100000] of longint;
procedure gj1;//高精乘
var
j:longint;
begin
for j:=1 to l do g[j]:=g[j]*b[i];
for j:=1 to l do
begin
g[j+1]:=g[j+1]+g[j] div 10;
g[j]:=g[j] mod 10;
end;
inc(l);
while g[l]>9 do
begin
g[l+1]:=g[l+1]+g[l] div 10;
g[l]:=g[l] mod 10;
inc(l);
end;
if g[l]=0 then dec(l);
end;
procedure gj2;//高精除
var
j:longint;
begin
for j:=l downto 1 do
begin
g[j-1]:=g[j-1]+(g[j] mod c[n])*10;
g[j]:=g[j] div c[n];
end;
while g[l]=0 do dec(l);
end;
procedure sort(l,r:longint);
var
i,j,x,y:longint;
begin
i:=l;
j:=r;
x:=a[(l+r) div 2];
repeat
while a[i]<x do inc(i);
while x<a[j] do dec(j);
if not (i>j) then
begin
y:=a[i];
a[i]:=a[j];
a[j]:=y;
y:=b[i];
b[i]:=b[j];
b[j]:=y;
y:=c[i];
c[i]:=c[j];
c[j]:=y;
inc(i);
dec(j);
end;
until (i>j);
if l<j then sort(l,j);
if i<r then sort(i,r);
end;
begin
readln(n);
readln(b[0],c[0]);
for i:=1 to n do
begin
read(b[i],c[i]);
a[i]:=b[i]*c[i];
end;
sort(1,n);
l:=1;
g[1]:=b[0];
for i:=1 to n-1 do gj1;
gj2;
for i:=l downto 1 do write(g[i]);
writeln;
end.

Sofa

program game;

uses
    math;

const
    size = 10000;

type
    hugeint = record
        len: integer;
        num: array [1..size] of longint;
    end;
    arr = array [0..1000] of integer;

var
    n: integer;
    a, b: arr;
    i, j: integer;
    c, d: longint;
    big, z, cal: hugeint;

function get(a: integer): integer;
var
    s: string;
    ans: integer;
begin
    str(a, s);
    ans := length(s);
    exit(ans);
end;

procedure geth(var a: hugeint);
var
    i: longint;
begin
    for i := 1 to size do
        if a.num[i] > 0 then
            a.len := i;
end;

function transh(a: longint): hugeint;
var
    i: integer;
    s: string;
    ans: hugeint;
begin
    str(a, s);
    fillchar(ans, sizeof(ans), 0);
    ans.len := length(s);
    for i := ans.len downto 1 do
        ans.num[i] := ord(s[ans.len - i + 1]) - 48;
    exit(ans);
end;

function over(a, b: hugeint): boolean;
var
    i: longint;
begin
    if (a.len < b.len) then
        exit(false);
    if (a.len > b.len) then
        exit(true);
    for i := a.len downto 1 do
    begin
        if a.num[i] < b.num[i] then
            exit(false);
        if a.num[i] > b.num[i] then
            exit(true);
    end;
    exit(true);
end;

function times(a: hugeint; b: integer): hugeint;
var
    i, j: longint;
    ans: hugeint;
begin
    fillchar(ans, sizeof(ans), 0);
    for i := 1 to a.len do
        ans.num[i] := a.num[i] * b;
    for i := 1 to (a.len + get(b) - 1) do
    begin
        inc(ans.num[i + 1], ans.num[i] div 10);
        ans.num[i] := ans.num[i] mod 10;
    end;
    geth(ans);
    exit(ans);
end;

function devide(a: hugeint; b: integer): hugeint;
var
    i, j, d: longint;
    ans: hugeint;
begin
    fillchar(ans, sizeof(ans), 0);
    d := 0;
    for i := a.len downto 1 do
    begin
        d := d * 10 + a.num[i];
        while d >= b do
        begin
            ans.num[i] := d div b;
            d := d mod b;
        end;
    end;
    geth(ans);
    exit(ans);
end;

procedure print(a: hugeint);
var
    i: longint;
begin
    for i := a.len downto 1 do
        write(a.num[i]);
end;

procedure qsort(var a, b: arr; left, right: integer);
var
    i, j: integer;
    k, temp: longint;
begin
    if (left >= right) then exit;
    i := left;
    j := right;
    k := a[i] * b[i];
    while (i <> j) do
    begin
        while (a[j] * b[j] >= k) and (j>i) do dec(j);
        while (k >= a[i] * b[i]) and (i<j) do inc(i);
        if (i < j) then
        begin
            temp := a[i]; a[i] := a[j]; a[j] := temp;
            temp := b[i]; b[i] := b[j]; b[j] := temp;
        end;
    end;
    temp := a[left]; a[left] := a[i]; a[i] := temp;
    temp := b[left]; b[left] := b[i]; b[i] := temp;
    qsort(a, b, left, i - 1);
    qsort(a, b, i + 1, right);
end;

begin
//    assign(input,'game.in'); assign(output,'game.out');
//    reset(input); rewrite(output);
    read(n);
    for i := 0 to n do read(a[i], b[i]);
    qsort(a, b, 1, n);
    fillchar(big, sizeof(big), 0);
    fillchar(z,sizeof(z), 0);
    z := transh(a[0]);
    for i := 1 to n do
    begin
        cal := devide(z, b[i]);
        z := times(z, a[i]);
        if over(cal, big) then
            big := cal;
    end;
    print(big);
//    close(input); close(output);
end.

这道题考的是RP啊= =看得出贪心了就不一样了。。不然只能乖乖算..