335 条题解
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9
Dragon___Killer LV 8 @ 2017-08-16 21:13:26
#py大法好 a = int(input()) b = int(input()) print(a*b) -
7
Sky1231 (sky1231) LV 10 @ 2017-10-04 17:53:44
#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; struct bigint { vector<int> num; int operator == (bigint b) { if (num.size()!=b.num.size()) return 0; else { for (unsigned long long i=0,size=num.size();i<size;i++) if (num[i]!=b.num[i]) return 0; return 1; } } int operator != (bigint b) { return ((!(*this==b))?1:0); } int operator < (bigint b) { if (num.size()<b.num.size()) return 1; else if (num.size()>b.num.size()) return 0; else { for (unsigned long long i=0,size=num.size();i<size;i++) { if (num[size-1-i]<b.num[size-1-i]) return 1; else if (num[size-1-i]>b.num[size-1-i]) return 0; } return 0; } } int operator > (bigint b) { if (num.size()<b.num.size()) return 0; else if (num.size()>b.num.size()) return 1; else { for (unsigned long long i=0,size=num.size();i<size;i++) { if (num[size-1-i]<b.num[size-1-i]) return 0; else if (num[size-1-i]>b.num[size-1-i]) return 1; } return 0; } } int operator <= (bigint b) { return ((!(*this>b))?1:0); } int operator >= (bigint b) { return ((!(*this<b))?1:0); } bigint mov(long long x) { bigint ans; ans.num.clear(); if (x+num.size()>0) { ans.num.resize(num.size()+x,0); for (long long i=0,size=num.size();i<size;i++) if (i+x>=0) ans.num[i+x]=num[i]; } else ans=0; return ans; } bigint operator = (long long b) { char s[20+1]; sprintf(s,"%lld",b); num.clear(); num.resize(strlen(s)); for (unsigned long long i=0,size=num.size();i<size;i++) num[i]=(s[size-1-i]-'0'); return (*this); } bigint operator = (string b) { num.clear(); num.resize(b.size()); for (unsigned long long i=0,size=num.size();i<size;i++) num[i]=(b[size-1-i]-'0'); return (*this); } bigint operator + (bigint b) { bigint ans; ans.num.clear(); ans.num.resize(max(num.size(),b.num.size())); for (unsigned long long i=0,size=ans.num.size();i<size;i++) ans.num[i]=((i<num.size())?num[i]:0)+((i<b.num.size())?b.num[i]:0); for (unsigned long long i=0,size=ans.num.size();i<size;i++) { if (ans.num[i]>=10&&i==size-1) ans.num.resize(++size); ans.num[i+1]+=(ans.num[i]/10); ans.num[i]%=10; } return ans; } bigint operator += (bigint b) { return (*this=(*this+b)); } bigint operator - (bigint b) { bigint ans; if ((*this)==b) { ans=0; return ans; } ans.num.clear(); ans.num.resize(max(num.size(),b.num.size())); for (unsigned long long i=0,size=ans.num.size();i<size;i++) ans.num[i]=((i<num.size())?num[i]:0)-((i<b.num.size())?b.num[i]:0); for (unsigned long long i=0,size=ans.num.size();i<size;i++) while (ans.num[i]<0) ans.num[i]+=10,ans.num[i+1]--; while (ans.num[ans.num.size()-1]==0) ans.num.resize(ans.num.size()-1); return ans; } bigint operator -= (bigint b) { return (*this=(*this-b)); } bigint operator * (bigint b) { bigint ans,num_0; num_0=0; ans.num.clear(); if (*this!=num_0&&b!=num_0) { ans.num.resize(num.size()+b.num.size()-1); for (unsigned long long i=0,size_1=num.size();i<size_1;i++) for (unsigned long long j=0,size_2=b.num.size();j<size_2;j++) ans.num[i+j]+=(num[i]*b.num[j]); for (unsigned long long i=0,size=ans.num.size();i<size;i++) { if (ans.num[i]>=10&&i==size-1) ans.num.resize(++size); ans.num[i+1]+=(ans.num[i]/10); ans.num[i]%=10; } } else ans=0; return ans; } bigint operator *= (bigint b) { return (*this=(*this*b)); } bigint operator / (bigint b) { bigint ans,num_0; num_0=0; if (*this<b) return num_0; if (*this!=num_0&&*this>num_0) { bigint x,y; x=*this,y=b.mov(num.size()-b.num.size()); ans.num.resize(num.size()-b.num.size()+1,0); for (long long i=ans.num.size()-1;i>=0;i--,y=y.mov(-1)) while (x>=y) x-=y,ans.num[i]++; for (unsigned long long i=0,size=ans.num.size();i<size;i++) { if (ans.num[i]>=10&&i==size-1) ans.num.resize(++size); ans.num[i+1]+=(ans.num[i]/10); ans.num[i]%=10; } while (ans.num.size()>1&&ans.num[ans.num.size()-1]==0) ans.num.resize(ans.num.size()-1); } else ans=0; return ans; } bigint operator /= (bigint b) { return (*this=(*this/b)); } void print() { for (unsigned long long i=0,size=num.size();i<size;i++) printf("%d",num[size-1-i]); } }; char s[10000+1]; #define s(x) s[x-1] bigint a,b,c; string s_a,s_b; #define s_a(x) s_a[x-1] #define s_b(x) s_b[x-1] int main() { while (~scanf("%s",s)) { s_a=s; getchar(); scanf("%s",s); s_b=s; getchar(); a=s_a,b=s_b; c=a*b; c.print(); printf("\n"); } }-
很H2O的題
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Spencer Moore @ 2017-12-28 09:39:59
DALAO
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ttbr5145 @ 2018-11-09 21:49:47
@sky1231: 大整数类？
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1
lk983773710 LV 7 @ 2018-07-04 00:00:44
#include <iostream>
#include <cstring>
using namespace std;
int a[10601],b[10601],c[20601];
int main(int argc, char** argv)
{
int la,lb,x=0,o;
string A,B;
cin>>A>>B;
la=A.length();
lb=B.length();
for(int i=1;i<=la;i++)
{
a[i]=A[la-i]-'0';
}
for(int j=1;j<=lb;j++)
{
b[j]=B[lb-j]-'0';
}
for(int k=1;k<=la;k++)
{
for(int l=1;l<=lb;l++)
{
c[k+l-1]=a[k]*b[l]+c[k+l-1]+x;
x=c[k+l-1]/10;
c[k+l-1]=c[k+l-1]%10;
}
c[k+lb]=x;
x=0;
}
o=la+lb;
while(c[o]==0&&o!=0)
{
o--;
}
for(int oo=o;oo>=1;oo--)
{
cout<<c[oo];
}
return 0;
} -
1
fan672867217 LV 6 @ 2018-03-18 10:01:31
这题用Java无敌
import java.math.BigInteger;
import java.util.Scanner;public class Main
{
public static void main(String[] args)
{
Scanner scan=new Scanner(System.in);
BigInteger a=scan.nextBigInteger();
BigInteger b=scan.nextBigInteger();
BigInteger c=a.multiply(b);
System.out.println(c);
}}
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1
Bang_Dream_ LV 4 @ 2017-12-11 22:02:11
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
string tp;struct u{
int A[10000],len;
//构造函数 指的是 一开始变量生成的时候做的初始化操作
u(){
memset(A,0,sizeof(A));
len=1;
}u (string a){
memset(A,0,sizeof(A));
this->len=0;
for(int i=a.length()-1;i>=0;i--){
this->A[this->len]=a[i]-'0';
this->len++;
}
}
void pt(){
for(int i=this->len-1;i>=0;i--){
printf("%d",A[i]);
}
//printf("\n%d",len);
}u operator * (u s)//重载运算符 *
{
u ans;
int maxlen=this->len+s.len+1;
for(int i=0;i<s.len;i++){
for(int j=0;j<this->len;j++){
ans.A[i+j]+=this->A[j]*s.A[i];
ans.A[i+j+1]+=ans.A[i+j]/10;
ans.A[i+j]%=10;
}
}
while(ans.A[maxlen]==0 && maxlen>0) maxlen--;
ans.len=maxlen+1;
return ans;
}};
int main(){
cin>>tp;
u a(tp);
cin>>tp;
u b(tp);
u c=a*b;
c.pt();
return 0;
}-
hh
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1
QAQ
#include<cstdio> #include<cstring> #include<algorithm> #include<iostream> #include<queue> #include<cstdlib> #include<cmath> #define maxn 100005 using namespace std; char s1[maxn],s2[maxn]; int c[maxn],a1[maxn],a2[maxn]; void mul(int a[],int b[]){ for(int i=1;i<=a[0];i++){ for(int j=i;j<=b[0]+i-1;j++){ c[j]+=a[i]*b[j-i+1]; if(c[j]>=10){ c[j+1]+=c[j]/10; c[j]=c[j]%10; } } }c[0]=a[0]+b[0]; while(c[c[0]]==0&&c[0]>0){c[0]--; } while(c[c[0]+1]){ c[0]++; c[c[0]+1]+=c[c[0]]/10; c[c[0]]%=10; } } int main(){ scanf("%s%s",s1+1,s2+1); int len1=strlen(s1+1),len2=strlen(s2+1); for(int i=1;i<=len1;i++){ a1[i]=s1[len1-i+1]-'0'; } for(int i=1;i<=len2;i++){ a2[i]=s2[len2-i+1]-'0'; }a1[0]=len1,a2[0]=len2; mul(a1,a2); for(int i=c[0];i>=1;i--){ printf("%d",c[i]); } }-
0的情况有错啊大佬
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1
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
#include <algorithm>
#include <set>
#include <map>
#include <bitset>
#include <cmath>
#include <queue>
#include <stack>
using namespace std;int main()
{
char a[10000] , b[10000];
int n[10000]={} , m[10000]={} , result[10000] = {};
int la , lb , i , j , w = 0;
int k = w;
cin >> a >> b;
la = strlen(a);
lb = strlen(b);
for(i = 0;i < la;i++) n[i] = a[la-i-1] - '0';
for(i = 0;i < lb;i++) m[i] = b[lb-i-1] - '0';
for(i = 0;i < la;i++)
for(j = 0;j < lb;j++)
{
result[i + j] += n[i] * m[j];
if(result[i + j] != 0) w = i + j;
}
for(i = 0;i <= w + 5;i++)
{
result[i + 1] += result[i] / 10;
result[i] %= 10;
if(result[i] != 0) k = i;
}
for(i = k;i >= 0;i--) cout << result[i];
return 0;
} -
1
好短的主函数
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
char cd[100000];
void maint(char ak[10001],char bk[10001])
{
int a[100000],b[100000],c[100001],d,e,f,z=0;
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
memset(c,0,sizeof(c));
d=strlen(ak);
e=strlen(bk);
for(int i=0;i<=d-1;++i)
{
a[d-i]=ak[i]-48;
}
for(int j=0;j<=e-1;++j)
{
b[e-j]=bk[j]-48;
}
for(int q=1;q<=d;++q)
{
z=0;
for(int v=1;v<=e;++v)
{
c[q+v-1]=a[q]*b[v]+z+c[q+v-1];
z=c[q+v-1]/10;
c[q+v-1]=c[q+v-1]%10;
}
c[q+e]=z;
}
f=d+e;
while((c[f]==0)&&(f>1))
{
f--;
}
for(int u=1;u<=f;++u)
{
cd[u]=0;
}
for(int u=f;u>=1;u--)
{
cd[f-u]=c[u]+48;
}
cout<<cd;
}
int main()
{
char a[10001],b[10001];
cin>>a>>b;
maint(a,b);
} -
1
#include <cstdio> #include <string> #include <algorithm> #include <iostream> using namespace std; #define MAX 1 // D[i] : 0....99999 int pow_int(int a,int n) { int count=1,x=a; while(n>0) { if(n&1) count*=x; x=x*x; n=n>>1; } return count; } struct BigInt { int D[100000]; int max; int n; BigInt(int x) { fill(D,D+100,0); int i; max=pow_int(10,MAX); for(i=0;x>0;i++) { D[i]=x%max; x=x/max; } n=i; } BigInt(void) { fill(D,D+100,0); n=0; max=pow_int(10,MAX); } void print(void) { cout<<D[n-1]; for(int i=n-2;i>=0;i--) { cout<<D[i]; } cout<<endl; } }; BigInt &mul(BigInt &a,BigInt &b); int main() { string s1,s2; cin>>s1>>s2; BigInt a,b; for(int i=s1.size()-1,j=0;i>=0;i--,j++) a.D[j]=s1[i]-'0'; a.n=s1.size(); for(int i=s2.size()-1,j=0;i>=0;i--,j++) b.D[j]=s2[i]-'0'; b.n=s2.size(); BigInt result=mul(a,b); result.print(); return 0; } BigInt &mul(BigInt &a,BigInt &b) { BigInt *result=new BigInt(0); for(int i=0;i<a.n;i++) { for(int j=0;j<b.n;j++) { result->D[i+j]+=a.D[i]*b.D[j]; if(result->D[i+j]>=result->max) result->D[i+j+1]+=result->D[i+j]/result->max; result->D[i+j]=result->D[i+j]%result->max; } } result->n=a.n+b.n-1; while(result->D[result->n]==0&&result->n>1) { result->n--; } if(result->D[result->n]) result->n++; return *result; } -
1
2351599120 LV 9 @ 2017-07-22 11:30:45
#include <iostream>
#include <iomanip>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <cctype>
#include <vector>
#include <queue>
#include <set>
#include <bitset>
#include <cassert>
#include <map>
#include <string>
#include <sstream>
#include <ctime>
using namespace std;
int a[20010],b[20010],c[20010];
char s1[20010],s2[20010],s3[20010];
int main()
{
int la,lb,l;
scanf("%s%s",s1,s2);
la=strlen(s1);
lb=strlen(s2);
for(int i=1;i<=la;i++)
{
a[i]=s1[la-i]-'0';
}
for(int i=1;i<=lb;i++)
{
b[i]=s2[lb-i]-'0';
}
for(int i=1;i<=la;i++)
{
for(int j=1;j<=lb;j++)
{
c[i+j-1]=c[i+j-1]+a[i]*b[j];
c[i+j]=c[i+j]+c[i+j-1]/10;
c[i+j-1]=c[i+j-1]%10;
}
}
l=la+lb;
while(c[l]==0&&l>1)
{
l--;
}
for(int i=l;i>=1;i--)
{
printf("%d",c[i]);
}
return 0;
}-
CatWigle281 @ 2020-07-18 19:13:04
望丰展？使MD
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1
蒟蒻写的非常简单的代码，并没有用FFT或者压位什么的
但是居然AC了......
供参考
c++
#include <cstdio>
#include <cstring>
using namespace std;
int a[10001],b[10001],c[20001];
char s1[10001],s2[10001];
int main() {
gets(s1);gets(s2);
int i,j,l1=strlen(s1),l2=strlen(s2),x=0;
for (i=1;i<=l1;i++) a[i]=s1[l1-i]-48;
for (i=1;i<=l2;i++) b[i]=s2[l2-i]-48;
for (i=1;i<=l1;i++)
for (j=1;j<=l2;j++)
c[i+j]+=a[i]*b[j];
for (i=1;i<=l1+l2;i++) { //统一整理进位
c[i+1]+=c[i]/10;
c[i]%=10;
}
int max=l1+l2+1;
while (c[max]==0) max--; //去除数字前的零
for (i=max;i>=2;i--) printf("%d",c[i]);
printf("\n");
return 0;
}
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0
简单易懂的写法，好吧
#include<bits/stdc++.h>
using namespace std;
int b[10000],c[10000],d[20000];
int main(){
string n,a;
cin>>n>>a;
for(int i=0;i<n.size();i++){
b[n.size()-i-1]=n[i]-'0';
}for(int i=0;i<a.size();i++){
c[a.size()-i-1]=a[i]-'0';
}
for(int i=0;i<n.size();i++){
for(int j=0;j<a.size();j++){
d[i+j]+=b[i]*c[j];
if(d[i+j]>=10){
d[i+j+1]=d[i+j]/10+d[i+j+1];
d[i+j]%=10;
}
}
}int q=n.size()+a.size();
int t=0;
for(int i=q;i>=0;i--){
if(d[i]!=0){
t=i;
break;
}
}for(int i=t;i>=0;i--){
cout<<d[i];
}return 0;
} -
0
#include<bits/stdc++.h> #define ll long long int using namespace std; struct bigint{ int a[7000],n,N=1000,t=3; bigint (){ memset(a,0,sizeof(a)); } bigint(char* s,int _n){ memset(a,0,sizeof(a)); char* _end=s+_n,*beg; int p=0; while(_end-t>s){ beg=_end-t; for(int i=0;i<t;i++){ a[p]*=10; a[p]+=beg[i]-'0'; } ++p; _end=beg; } if(_end>s){ for(int i=0;i<_end-s;i++){ a[p]*=10; a[p]+=s[i]-'0'; } ++p; } n=p; } int& operator [](int i){ return a[i]; } bigint operator * (bigint& b){ int i,j,m=b.n; bigint r; int sr=n+m-1; for(i=0;i<n;++i) for(j=0;j<m;++j) r[i+j]+=a[i]*b[j]; for(i=0;i<sr;++i){ r[i+1]+=r[i]/N; r[i]%=N; } if(r[sr])sr++; r.n=sr; return r; } void show(){ int i=n-2; printf("%d",a[n-1]); while(i>=0) printf("%03d",a[i--]); } }; char s[10001]; int main() { //freopen("in.txt","r",stdin); scanf("%s",s); bigint a(s,strlen(s)); scanf("%s",s); bigint b(s,strlen(s)); (a*b).show(); }后两个wa了的，可能是中间爆int了，压四位先乘了再进位会爆int...要不用longlong 要不就压三位
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0
#include<iostream>
#include<string>
using namespace std;
const int N = 10001;
void h_multiply_2(int a[], int b[], int c[])
{
int la = a[0];
int lb = b[0];
for(int i=1;i<=la;i++)
{
for(int j=1;j<=lb;j++)
{
c[i+j-1] += a[i] * b[j];
}
}
for(int i=1;i<=la+lb;i++)
{
c[i+1] += c[i] / 10;
c[i] %= 10;
}
c[0] = la+lb;
}
int convert(string s,int n[])
{
int length = s.length();
for(int i=1;i<=length;i++)
{
n[i] = s[length-i] - '0';
}
n[0] = length;
return length;
}
void output(int n[])//n是算好了的高精度数
{
int length = n[0];
if(length==0)
{
cout << 0 << endl;
return;
}
for(int i=length;i>=1;i--)
{
cout << n[i];
}
cout << endl;
}
int trim(int n[])
{
int length = 0;
for(int i=n[0];i>0;i--)
{
if(n[i]!=0)
{
length = i;
break;
}
}
n[0] = length;
return length;
}
int a[N+1];
int b[N+1];
int c2[2*N+1];
int main()
{
string s1, s2;
cin >> s1 >> s2;
convert(s1, a);
convert(s2, b);
h_multiply_2(a,b,c2);
trim(c2);
output(c2);
return 0；
} -
0
此人太弱，用户名无法显示 (蒟蒻在此-QwQ-) LV 7 @ 2020-07-12 14:05:39
x=int(input()) y=int(input()) print(x*y)真香！！
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0
告诉你什么叫做流！弊！
Python3三行代码解决：x=int(input()) #input()用于读入一行字符，int()用于把字串转为数字 y=int(input());#在python中分号可以省略，也可以加上 print(x*y) #输出a*b，python中有乘方运算符（^）哇哈哈哈哈哈哈哈哈哈哈哈哈~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
//笑出猪叫
Dragon___Killer
LV 8
@ 2017-08-16 21:13:26
