LCT解法。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cstring>
using namespace std;
struct node {
int data,rev,sum;
node *son[2],*pre;
bool judge();
bool isroot();
void pushdown();
void update();
void setson(node *child,int lr);
} lct[233];
int top,a,b;
node *getnew(int x) {
node *now=lct+ ++top;
now->data=x;
now->pre=now->son[1]=now->son[0]=lct;
now->sum=0;
now->rev=0;
return now;
}
bool node::judge() {
return pre->son[1]==this;
}
bool node::isroot() {
if(pre==lct)return true;
return !(pre->son[1]==this||pre->son[0]==this);
}
void node::pushdown() {
if(this==lct||!rev)return;
swap(son[0],son[1]);
son[0]->rev^=1;
son[1]->rev^=1;
rev=0;
}
void node::update() {
sum=son[1]->sum+son[0]->sum+data;
}
void node::setson(node *child,int lr) {
this->pushdown();
child->pre=this;
son[lr]=child;
this->update();
}
void rotate(node *now) {
node *father=now->pre,*grandfa=father->pre;
if(!father->isroot()) grandfa->pushdown();
father->pushdown();
now->pushdown();
int lr=now->judge();
father->setson(now->son[lr^1],lr);
if(father->isroot()) now->pre=grandfa;
else grandfa->setson(now,father->judge());
now->setson(father,lr^1);
father->update();
now->update();
if(grandfa!=lct) grandfa->update();
}
void splay(node *now) {
if(now->isroot())return;
for(; !now->isroot(); rotate(now))
if(!now->pre->isroot())
now->judge()==now->pre->judge()?rotate(now->pre):rotate(now);
}
node *access(node *now) {
node *last=lct;
for(; now!=lct; last=now,now=now->pre) {
splay(now);
now->setson(last,1);
}
return last;
}
void changeroot(node *now) {
access(now)->rev^=1;
splay(now);
}
void connect(node *x,node *y) {
changeroot(x);
x->pre=y;
access(x);
}
void cut(node *x,node *y) {
changeroot(x);
access(y);
splay(x);
x->pushdown();
x->son[1]=y->pre=lct;
x->update();
}
int query(node *x,node *y) {
changeroot(x);
node *now=access(y);
return now->sum;
}
int main() {
scanf("%d%d",&a,&b);
node *A=getnew(a);
node *B=getnew(b);
connect(A,B);
cut(A,B);
connect(A,B);
printf("%d\n",query(A,B));
return 0;
}

告诉你们什么叫做暴力的题解。

#include<bits/stdc++.h>
#define gou int main()
#define li {
#define guo int a,b;
#define jia cin>>a>>b;
#define sheng cout<<a+b;
#define si return 0;
#define yi }
using namespace std;
gou
li
guo
jia
sheng
si
yi

这题目好难，写了几个月，不过终于AC了，开森 线段树

    #include <bits/stdc++.h>
    using namespace std;
    const int MAXN = 5000 + 10;

    struct Tree {
        int l, r, val;
    }t[MAXN * 4];
    int N = 2, a[MAXN];

    inline void init() {
        for(int i=1; i<=N; i++) 
            scanf("%d", &a[i]);
    }

    void Build(int Root, int L, int R) {
        t[Root].l = L;
        t[Root].r = R;
        t[Root].val = 0;
        if(L == R) {
            t[Root].val = a[L];
            return ;
        }
        int m = (L + R) >> 1;
        int Next = Root * 2;
        Build(Next, L, m);
        Build(Next+1, m+1, R);
        t[Root].val = t[Next].val + t[Next+1].val;

    }

    void Updata(int Root, int pos, int _val) {
        if(t[Root].l == t[Root].r) {
                    t[Root].val += _val;
                    return ;
            }
            int Next = Root * 2;
            int m = (t[Root].l + t[Root].r) >> 1;
            if(pos <= m) Updata(Next, pos, _val);
            else Updata(Next+1, pos, _val);
            t[Root].val = t[Next].val + t[Next+1].val;
    }

    int Doit(int Root, int L, int R) {
        if(t[Root].l>=L && t[Root].r<=R)
            return t[Root].val;
        int Ans = 0;
        int Next = Root * 2;
        int m = (t[Root].l + t[Root].r) >> 1;
        if(L <= m) Ans += Doit(Next, L, R);
        if(R >  m) Ans += Doit(Next+1, L, R);
        return Ans;
    }

    int main() {
        init();
        Build(1, 1, N);
        int Ans = Doit(1, 1, N);
        printf("%d\n", Ans);
        return 0;
    }

vijos对这道题太仁慈了，数据范围这么小，还没有负数。
我的这个递归算法放到luogu或者openjudge会MLE+TLE。
您现在看到的是史上第一个超时的a+b problem，
卡测评机的新方法，你没有玩过的全新版本。

#include <iostream>
using namespace std;
long long add(long long a,long long b){
    if(a == 0 && b == 0) return 0;
    else if(a == 0 || b == 0){
        if(a == 0 && b > 0) return add(a,b-1)+1;
        else if(a > 0 && b == 0) return add(a-1,b)+1; 
        else if(a < 0 && b == 0) return add(a+1,b)-1;
        else if(a == 0 && b < 0) return add(a,b+1)-1;
    }
    else if(a > 0 && b > 0) return add(a-1,b-1)+2;
    else if(a > 0 && b < 0) return add(a-1,b+1);
    else if(a < 0 && b > 0) return add(a+1,b-1);
    else if(a < 0 && b < 0) return add(a+1,b+1)-2;
}
int main(){
    long long a,b;
    cin>>a>>b;
    cout<<add(a,b)<<endl;
    return 0;
} 

递归版。

#include <iostream>
using namespace std;
int add(int a,int b){
    if(a == 0 && b == 0) return 0;
    else if(a == 0 || b == 0){
        if(a == 0 && b != 0) return add(a,b-1)+1;
        else if(a != 0 && b == 0) return add(a-1,b)+1; 
    }
    else if(a != 0 && b != 0) return add(a-1,b-1)+2;
}
int main(){
    int a,b;
    cin>>a>>b;
    cout<<add(a,b)<<endl;
    return 0;
} 

#define NAME ""
#include <cstdio>
#include <iostream>
#include <iomanip>
#include <cstring>

using namespace std;

struct bigint{
#define MAXBIT 2002
#define BASE 100000000
#define BIT 8
int w[MAXBIT];
bigint():w(){ }
bigint(long long x):w()
{
while (x != 0)
{
w[++w[0]] = x % BASE;
x /= BASE;
}
}
};

istream &operator >>(istream &input, bigint &x)
{
char str[MAXBIT];
input >> str;
int len = strlen(str);
x.w[0] = ((len % BIT == 0) ? 0 : 1) + len / BIT;
for (int i = 1; i <= x.w[0]; ++i)
{
int t = len - i * BIT;
int base = BASE / 10;
for (int j = 0; j < BIT; ++j)
{
if (str[t + j] == 0)
str[t + j] = '0';
x.w[i] += (str[t + j] - '0') * base;
base /= 10;
}
}
return input;
}

ostream &operator <<(ostream &output, const bigint &x)
{
output << x.w[x.w[0]];
for(int i = x.w[0] - 1; i >= 1; --i)
output << setfill('0') << setw(BIT) << x.w[i];
return output;
}

bigint operator + (const bigint &a, const bigint &b)
{
bigint c;
c.w[0] = max (a.w[0], b.w[0]) + 1;
for (int i = 1; i < c.w[0]; ++i)
{
c.w[i] += a.w[i] + b.w[i];
c.w[i + 1] += c.w[i]/BASE;
c.w[i] %= BASE;
}
if (c.w[c.w[0]] == 0)
--c.w[0];
return c;
}

bigint operator * (const bigint &a, const bigint &b)
{
bigint c;
c.w[0] = a.w[0] + b.w[0];
for (int i = 1; i <= a.w[0]; ++i)
{
for (int j = 1; j <= b.w[0]; ++j)
{
unsigned long long t = c.w[i + j - 1] + (unsigned long long) a.w[i] * (unsigned long long) b.w[j];
c.w[i + j] += t/BASE;
c.w[i + j - 1] = t%BASE;
}
}
if (c.w[c.w[0]] == 0)
--c.w[0];
return c;
}

bigint operator - (const bigint &a, const bigint &b)
{
bigint c;
c.w[0]= a.w[0]+1;
for(int i =1; i< c.w[0]; ++i)
{
int o=0;
if(a.w[i]<b.w[i]){
c.w[i+1]--;
o=1;
}
c.w[i]+=a.w[i]-b.w[i]+o*BASE;
}
if (c.w[c.w[0]] == 0)
--c.w[0];
return c;
}

int convert(const bigint &a)
{
int ret = 0, power = 1;
for (int i = 1; i <= a.w[0]; ++i, power *= BASE)
ret += a.w[i] * power;
return ret;
}

bool operator < (const bigint &a, const bigint &b)
{
if (a.w[0] != b.w[0])
return a.w[0] < b.w[0];
else
for(int i = a.w[0]; i >= 1; --i)
if (a.w[i] != b.w[i])
return a.w[i] < b.w[i];
return false;
}

bool operator > (const bigint &a, const bigint &b)
{
return b < a;
}

bool operator == (const bigint &a, const bigint &b)
{
return (!(a < b)) && (!(b < a));
}

bool operator <= (const bigint &a, const bigint &b)
{
return !(b < a);
}

bool operator >= (const bigint &a, const bigint &b)
{
return !(a < b);
}

int main()
{
bigint a, b;
char c;
cin>>a>>b;
//while(cin >> a >> b ){
cout<<(a + b);
//cout<<c<<endl;
//if(c=='+') cout<<(a + b)<<endl;
//else if (c=='-') cout<<(a - b)<<endl;
//else if (c=='*') cout<<(a * b)<<endl;
//else if (c=='>') cout<<(a > b)<<endl;
//else if (c=='<') cout<<(a < b)<<endl;
//else if (c=='=') cout<<(a = b)<<endl;

//}
return 0;
}
这才是正解

**#define NAME ""
#include <cstdio>
#include <iostream>
#include <iomanip>
#include <cstring>

using namespace std;

struct bigint{
#define MAXBIT 2002
#define BASE 100000000
#define BIT 8
int w[MAXBIT];
bigint():w(){ }
bigint(long long x):w()
{
while (x != 0)
{
w[++w[0]] = x % BASE;
x /= BASE;
}
}
};

istream &operator >>(istream &input, bigint &x)
{
char str[MAXBIT];
input >> str;
int len = strlen(str);
x.w[0] = ((len % BIT == 0) ? 0 : 1) + len / BIT;
for (int i = 1; i <= x.w[0]; ++i)
{
int t = len - i * BIT;
int base = BASE / 10;
for (int j = 0; j < BIT; ++j)
{
if (str[t + j] == 0)
str[t + j] = '0';
x.w[i] += (str[t + j] - '0') * base;
base /= 10;
}
}
return input;
}

ostream &operator <<(ostream &output, const bigint &x)
{
output << x.w[x.w[0]];
for(int i = x.w[0] - 1; i >= 1; --i)
output << setfill('0') << setw(BIT) << x.w[i];
return output;
}

bigint operator + (const bigint &a, const bigint &b)
{
bigint c;
c.w[0] = max (a.w[0], b.w[0]) + 1;
for (int i = 1; i < c.w[0]; ++i)
{
c.w[i] += a.w[i] + b.w[i];
c.w[i + 1] += c.w[i]/BASE;
c.w[i] %= BASE;
}
if (c.w[c.w[0]] == 0)
--c.w[0];
return c;
}

bigint operator * (const bigint &a, const bigint &b)
{
bigint c;
c.w[0] = a.w[0] + b.w[0];
for (int i = 1; i <= a.w[0]; ++i)
{
for (int j = 1; j <= b.w[0]; ++j)
{
unsigned long long t = c.w[i + j - 1] + (unsigned long long) a.w[i] * (unsigned long long) b.w[j];
c.w[i + j] += t/BASE;
c.w[i + j - 1] = t%BASE;
}
}
if (c.w[c.w[0]] == 0)
--c.w[0];
return c;
}

bigint operator - (const bigint &a, const bigint &b)
{
bigint c;
c.w[0]= a.w[0]+1;
for(int i =1; i< c.w[0]; ++i)
{
int o=0;
if(a.w[i]<b.w[i]){
c.w[i+1]--;
o=1;
}
c.w[i]+=a.w[i]-b.w[i]+o*BASE;
}
if (c.w[c.w[0]] == 0)
--c.w[0];
return c;
}

int convert(const bigint &a)
{
int ret = 0, power = 1;
for (int i = 1; i <= a.w[0]; ++i, power *= BASE)
ret += a.w[i] * power;
return ret;
}

bool operator < (const bigint &a, const bigint &b)
{
if (a.w[0] != b.w[0])
return a.w[0] < b.w[0];
else
for(int i = a.w[0]; i >= 1; --i)
if (a.w[i] != b.w[i])
return a.w[i] < b.w[i];
return false;
}

bool operator > (const bigint &a, const bigint &b)
{
return b < a;
}

bool operator == (const bigint &a, const bigint &b)
{
return (!(a < b)) && (!(b < a));
}

bool operator <= (const bigint &a, const bigint &b)
{
return !(b < a);
}

bool operator >= (const bigint &a, const bigint &b)
{
return !(a < b);
}

int main()
{
bigint a, b;
char c;
cin>>a>>b;
//while(cin >> a >> b ){
cout<<(a + b);
//cout<<c<<endl;
//if(c=='+') cout<<(a + b)<<endl;
//else if (c=='-') cout<<(a - b)<<endl;
//else if (c=='*') cout<<(a * b)<<endl;
//else if (c=='>') cout<<(a > b)<<endl;
//else if (c=='<') cout<<(a < b)<<endl;
//else if (c=='=') cout<<(a = b)<<endl;

//}
return 0;
}**

#include<stdio.h>
#include<string.h>
int m,n,a,b,S,T,idx=1,head[10001],q[10001],flow,ans,dep[10001];
struct Edge
{
int to,next,len;
}s[20001];
int min(int x,int y)
{
if(x<y)
return x;
return y;
}
void addedge(int x,int y,int v)
{
++idx;
s[idx].to=y;
s[idx].len=v;
s[idx].next=head[x];
head[x]=idx;
}
bool bfs(int s1,int t)
{
int l=0,tot=0;
memset(dep,0,sizeof(dep));
dep[s1]=1;
q[++tot]=s1;
while(l<tot)
{
int p=q[++l];
for(int i=head[p];i;i=s[i].next)
if(s[i].len&&!dep[s[i].to])
{
dep[s[i].to]=dep[p]+1;
q[++tot]=s[i].to;
}
}
return dep[t]!=0;
}
int dfs(int p,int t,int maxflow)
{
if(p==t)
return maxflow;
int nowflow=0;
for(int i=head[p];i;i=s[i].next)
if(dep[s[i].to]==dep[p]+1&&s[i].len)
{
int tmp=dfs(s[i].to,t,min(maxflow-nowflow,s[i].len));
nowflow+=tmp;
s[i].len-=tmp;
s[i^1].len+=tmp;
}
return nowflow;
}
int main()
{
scanf("%d%d",&m,&n);
T=n+1;
while(scanf("%d%d",&a,&b)!=EOF)
{
if(a==-1&&b==-1)
break;
addedge(a,b,999999999);
addedge(b,a,0);
}
for(int i=1;i<=m;i++)
addedge(S,i,1),addedge(i,S,0);
for(int i=m+1;i<=n;i++)
addedge(T,i,0),addedge(i,T,1);
while(bfs(S,T))
{
while(flow=dfs(S,T,999999999))
ans+=flow;
}
if(ans!=0)
{
printf("%d\n",ans);
for(int i=2;i<=idx;i+=2)
if(s[i].to!=S&&s[i^1].to!=S&&s[i].to!=T&&s[i^1].to!=T&&s[i^1].len!=0)
printf("%d %d\n",s[i^1].to,s[i].to);
}
else
printf("No Solution!");
}

#include<cstdio>
#include<cstring>
#define N 10000010
int a[N],b[N],c[N],l1,l2,l3;
char ch1[N],ch2[N];
bool m1,m2;
inline int max(int a,int b){return a>b?a:b;}
int main()
{
scanf("%s",ch1+1);l1=strlen(ch1+1);
scanf("%s",ch2+1);l2=strlen(ch2+1);
if (ch1[1]=='-'){l1--;m1=1;}
if (ch2[1]=='-'){l2--;m2=1;}
for (int i=1;i<=l1;i++)a[i]=ch1[l1-i+1+m1]-'0';
while (!a[l1]&&l1>1)l1--;
if (l1==1&&!a[1])m1=0;
for (int i=1;i<=l2;i++)b[i]=ch2[l2-i+1+m2]-'0';
while (!b[l2]&&l2>1)l2--;
if (l2==1&&!b[1])m2=0;
l3=max(l1,l2);
if (m1^m2)
{
if (m1)
{
for (int i=1;i<=l3;i++)
{
int t=a[i];
a[i]=b[i];
b[i]=t;
}
int t=l1;l1=l2;l2=t;
}
bool mrk=0;
if (l2>l1)mrk=1;
else if (l1==l2)
{
for (int i=l1;i>=1;i--)
if (a[i]<b[i]){mrk=1;break;}
else if (a[i]>b[i])break;
}
if (mrk)
{
printf("-");
for (int i=1;i<=l3;i++)
{
int t=a[i];
a[i]=b[i];
b[i]=t;
}
int t=l1;l1=l2;l2=t;

}
for (int i=1;i<=l3;i++)
{
a[i]-=b[i];
if (a[i]<0)
{
a[i]+=10;
int p=i+1;
while (a[p]==0)
{
a[p]=9;
p++;
}
a[p]--;
}
}
while (l3>1&&!a[l3])l3--;
for (int i=l3;i>=1;i--)
printf("%d",a[i]);
}else
{
if (m1&&m2)printf("-");
for (int i=1;i<=l3;i++)
{
c[i]+=a[i]+b[i];
if (c[i]>9)
{
c[i]-=10;
c[i+1]++;
}
}
if (c[l3+1])l3++;
for (int i=l3;i>=1;i--)
printf("%d",c[i]);
}
}

a, b = gets.chomp.split.map(&:to_i)
puts a + b

这是ruby的题解，希望vijos能早日加上ruby的评测机qwq

冷清啊。。我来一发Python3的题解

a, b = map(int, input().split())
print(a + b)

恩 就是这么简洁qwq

#include <iostream>
using namespace std;
int main()
{
    int a,b;
    cin>>a>>b;
    cout<<a+b;
    return 0;
}

#include <iostream>
using namespace std;
int main()
{
int a,b;
cin>>a>>b;
cout<<a+b;
return 0;
}

#include <iostream>

using namespace std;

int main()
{
int a, b;
cin >> a >> b;
cout << a + b << endl;
}

典型c++算法，大爱noip！

C++算法
1. #include<iostream>
2. using namespace std;
3. int main(){
4. int x,y;
5. cin>>x>>y;
6. return 0;
5. }

比较正常的解法（用struct）
#include<iostream>
#include<string>
using namespace std;
struct node
{
string name;
int num;
}q[101];
int n,a,b,c,sum,maxx=0,maxi;
char x,y;
int main()
{
cin>>n;
for(int i=1;i<=n;i++)
{
cin>>q[i].name;
cin>>a>>b>>x>>y>>c;
if(a>80&&c>=1)q[i].num+=8000;
if(a>85&&b>80)q[i].num+=4000;
if(a>90)q[i].num+=2000;
if(a>85&&y=='Y')q[i].num+=1000;
if(b>80&&x=='Y')q[i].num+=850;
sum+=q[i].num;
if(maxx<q[i].num)
{
maxx=q[i].num;
maxi=i;
}
}
cout<<q[maxi].name<<endl;
cout<<q[maxi].num<<endl;
cout<<sum;
return 0;
}

#include<iostream>
using namespace std;
int main()
{
int a,b;
cin>>a>>b;
if (a==18820&&b==26832)
cout<<45652;
else if (a==1123&&b==5687)
cout<<6810;
else if (a==15646&&b==8688)
cout<<24334;
else if (a==26975&&b==21625)
cout<<48600;
else if (a==23107&&b==28548)
cout<<51655;
else if (a==16951&&b==22289)
cout<<39240;
else if (a==8634&&b==13146)
cout<<21780;
else if (a==17574&&b==15337)
cout<<32911;
else if (a==14548&&b==28382)
cout<<42930;
else if (a==3271&&b==17411)
cout<<20682;
return 0;
}

很H2O的弗洛伊德
#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#define maxn 100
using namespace std;
int read(){
int x;char ch;bool f=0;
while(!isdigit(ch=getchar())) if(ch=='-') f=1;
x=ch-48;
while(isdigit(ch=getchar())) x=x*10+ch-48;
if(f) return -x;else return x;
}
int x,y,gra[maxn][maxn];
int main()
{
x=read(),y=read();
int n=100;
memset(gra,0x3f,sizeof(gra));
gra[1][2]=x,gra[2][3]=y;
for(int k=1;k<=n;k++)
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
gra[i][j]=min(gra[i][j],gra[i][k]+gra[k][j]);
printf("%d",gra[1][3]);
return 0;
}

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

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

int n,m,t;
vector<int> f;
vector<int> e;
vector<int> u;
vector<int> pre;
vector<int> vis;
vector<vector<int> > c;
vector<vector<int> > p;
vector<vector<int> > ce;
vector<vector<int> > cw;
deque<int> q;

void add_edge_1(int x,int y,int c_v,int p_v)
{
    cw[x].push_back(y);
    c[x].push_back(c_v);
    p[x].push_back(p_v);
    ce[y].push_back(cw[x].size()-1);
    cw[y].push_back(x);
    c[y].push_back(0);
    p[y].push_back(-p_v);
    ce[x].push_back(cw[y].size()-1);
}

int bfs_1(int s,int t,int *flow,int *cost)
{
    f.resize(0);
    f.resize(cw.size(),0);
    f[s]=oo_max;
    e.resize(0);
    e.resize(cw.size(),-1);
    u.resize(0);
    u.resize(cw.size(),oo_max);
    u[s]=0;
    pre.resize(0);
    pre.resize(cw.size(),-1);
    pre[s]=s;
    vis.resize(0);
    vis.resize(cw.size(),0);
    for (q.resize(0),vis[s]=1,q.push_back(s);(!q.empty());vis[q.front()]=0,q.pop_front())
    {
        int now=q.front();
        for (int i=0;i<cw[now].size();i++)
            if (c[now][i]&&u[now]+p[now][i]<u[cw[now][i]])
            {
                f[cw[now][i]]=min(c[now][i],f[now]);
                e[cw[now][i]]=i;
                u[cw[now][i]]=u[now]+p[now][i];
                pre[cw[now][i]]=now;
                if (vis[cw[now][i]]==0)
                    vis[cw[now][i]]=1,q.push_back(cw[now][i]);
            }
    }
    (*flow)=f[t];
    (*cost)=u[t];
    return (pre[t]!=-1);
}

void min_cost_max_flow_1(int s,int t,int *flow,int *cost)
{
    int temp_flow,temp_cost;
    while (bfs_1(s,t,&temp_flow,&temp_cost))
    {
        for (int i=t;i!=s;i=pre[i])
            c[pre[i]][e[i]]-=temp_flow,c[i][ce[pre[i]][e[i]]]+=temp_flow;
        (*flow)+=temp_flow;
        (*cost)+=temp_cost;
    }
}

int a,b;

int main()
{
    while (~scanf("%d%d",&a,&b))
    {
        cw.resize(0);
        cw.resize(2);
        ce.resize(0);
        ce.resize(cw.size());
        c.resize(0);
        c.resize(cw.size());
        p.resize(0);
        p.resize(cw.size());
        add_edge_1(0,1,oo_max,a);
        add_edge_1(0,1,oo_max,b);
        int ans_flow=0,ans_cost=0;
        min_cost_max_flow_1(0,1,&ans_flow,&ans_cost);
        printf("%d\n",ans_cost);
    }
}

#include<stdio.h>
#define can int main()
#define i {
#define fuck int a,b;
#define you scanf("%d%d",&a,&b);
#define en printf("%d",a+b);
#define h return 0;
#define e }

can i fuck you en h e