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

#include <stdio.h>
#include <windows.h>
#include <time.h>
int main () {
    int a, b;
    scanf("%d %d", &a, &b);
    srand(time(NULL));
    for (;;) {
        int c = rand() % (a + b) + 1;
        if (c == a + b) {
            printf("%d", c);
            return 0;       
        }
    }
}

这题去年IOI最难的，正解随机数，无他解，其他的都是牛马题解！！！！
虽然这是正解，但还是有可能TLE！！！！

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

上次的题解有点问题，但是通过我们IOI中国队120948572341241294812个夜晚的思考，终于研究出了该世纪难题的真正的正解，加了一些优化，已经可以攻破很多个点的TLE了，但仍有待改进
CODE:

#include <stdio.h>
#include <windows.h>
#include <time.h>

#define N 100000
#define o3 __attribute__((optimize("-O3")))
#define r register

#pragma GCC optimize(3)
#pragma GCC optimize(2)
#pragma GCC optimize("-Ofast")

o3 int main () {
    int a, b, before[N], l = 0;
    bool bz = true;
    scanf("%d %d", &a, &b);
    srand(time(NULL));
    for (;;) {
        int c = rand() % (a + b) + 1;
        for (r int i = 1; i <= l; ++ i) {
            if (before[i] == c) {
                bz = false;
                break;
            }
        }
        if (bz == false) {
            bz = true;
            continue;
        }
        if (c == a + b and bz == true) {
            printf("%d", c);
            before[++ l] = c;
            return 0;       
        }
    }
}

希望有能力的牛马可以帮忙优化一下！！！！！！

*

*#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;
}*

*

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

这题是Vijos的经典啊……

这里给出一些比较基础的A+B方法

SPFA：

#include<cstdio>
using namespace std;
int n,m,a,b,op,head[200009],next[200009],dis[200009],len[200009],v[200009],l,r,team[200009],pd[100009],u,v1,e;
int lt(int x,int y,int z)
{
    op++,v[op]=y;
    next[op]=head[x],head[x]=op,len[op]=z;
}
int SPFA(int s,int f)//SPFA……
{
    for(int i=1;i<=200009;i++){dis[i]=999999999;}
    l=0,r=1,team[1]=s,pd[s]=1,dis[s]=0;
    while(l!=r)
    {
        l=(l+1)%90000,u=team[l],pd[u]=0,e=head[u];
        while(e!=0)
        {
            v1=v[e];
            if(dis[v1]>dis[u]+len[e])
            {
                dis[v1]=dis[u]+len[e];
                if(!pd[v1])
                {
                    r=(r+1)%90000,
                    team[r]=v1,
                    pd[v1]=1;
                }
            }
            e=next[e];
        } 
    }
    return dis[f];
}
int main()
{
    scanf("%d%d",&a,&b);
    lt(1,2,a);lt(2,3,b);//1到2为a，2到3为b，1到3即为a+b……
    printf("%d",SPFA(1,3));
    return 0;
}

Floyd：

#include<iostream>
#include<cstring>
using namespace std;
long long n=3,a,b,dis[4][4];
int main()
{
    cin>>a>>b;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
        {
            dis[i][j]=2147483647;
        }
    }
    dis[1][2]=a,dis[2][3]=b;
    for(int k=1;k<=n;k++)
    {
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]);//Floyd……
            }
        }
    }
    cout<<dis[1][3];
}

递归：

#include<iostream>
using namespace std;
long long a,b,c;
long long dg(long long a)
{
    if(a<=5){return a;}//防超时……
    return (dg(a/2)+dg(a-a/2));
}
int main()
{
    cin>>a>>b;
    c=dg(a)+dg(b);
    cout<<c;
}

高精：

#include<iostream>
#include<cstring>
using namespace std;
int main()
{
    char a1[1000],b1[1000];
      int a[1000]={0},b[1000]={0},c[1000]={0},la,lb,lc,i,x;
      cin>>a1>>b1;
      la=strlen(a1);
      lb=strlen(b1);
      for(i=0;i<=la-1;i++){a[la-i]=a1[i]-48;}
    for(i=0;i<=lb-1;i++){b[lb-i]=b1[i]-48;}
      lc=1,x=0;
    while(lc<=la||lc<=lb){c[lc]=a[lc]+b[lc]+x,x=c[lc]/10,c[lc]%=10,lc++;}
    c[lc]=x;
    if(c[lc]==0){lc--;}
    for(i=lc;i>=1;i--){cout<<c[i];}
    cout<<endl;
    return 0;
}

压位高精：

#include <cstdio>  
#include <cstring>  
#include <cstdlib>  
#include <iostream>  
#define p 8
#define carry 100000000
using namespace std;  
const int Maxn=50001;  
char s1[Maxn],s2[Maxn];  
int a[Maxn],b[Maxn],ans[Maxn];  
int change(char s[],int n[])   
{  
    char temp[Maxn];   
    int len=strlen(s+1),cur=0;  
    while(len/p)
    {  
        strncpy(temp,s+len-p+1,p);
        n[++cur]=atoi(temp); 
        len-=p;
    }  
    if(len)
    {
        memset(temp,0,sizeof(temp));  
        strncpy(temp,s+1,len);  
        n[++cur]=atoi(temp);   
    }  
    return cur;
}  
int add(int a[],int b[],int c[],int l1,int l2)  
{  
    int x=0,l3=max(l1,l2);  
    for(int i=1;i<=l3;i++)
    {  
        c[i]=a[i]+b[i]+x;  
        x=c[i]/carry;
        c[i]%=carry;  
    }  
    while(x>0){c[++l3]=x%10;x/=10;}  
    return l3;
}  
void print(int a[],int len)  
{   
    printf("%d",a[len]);
    for(int i=len-1;i>=1;i--)printf("%0*d",p,a[i]);
    printf("\n");  
}  
int main()  
{
    scanf("%s%s",s1+1,s2+1);
    int la=change(s1,a);
    int lb=change(s2,b);
    int len=add(a,b,ans,la,lb);    
    print(ans,len);
}  

#include <iostream>

using namespace std;

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

最简单的题？按题目模拟即可

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
int a, b;
int main()
{
    scanf("%d %d", &a, &b);
    printf("%d", a + b);
    return 0;
}

#include <bits/stdc++.h>

int main(int a, int b, int k)
{
    if (k) scanf("%d%d", &a, &b);
    b == 0 ? printf("%d\n", a) : main(a ^ b, (a & b) << 1, 0);
}

main递归 + 模拟位运算

216
4165

#include<bits/stdc++.h>
using namespace std;
int main()
{
int a,b;
cin>>a>>b;
cout<<a+b;
return 0;

}

作死ing

#include<bits/stdc++.h>
using namespace std;
char s[10001],ss[10001];
int a[10001],b[10001],c[10001],j;
bool x=false;
int main() {
    memset(a,0,sizeof(a));
    memset(b,0,sizeof(b));
    memset(c,0,sizeof(c));
    scanf("%s%s",s,ss);
    a[0]=strlen(s);
    b[0]=strlen(ss);
    for(int i=1; i<=a[0]; i++) a[i]=s[a[0]-i]-'0';
    for(int i=1; i<=b[0]; i++) b[i]=ss[b[0]-i]-'0';
    for(j=1; j<=max(a[0],b[0])+1; j++) {
        c[j]=a[j]+b[j];
        if(c[j]>=10) {
            c[j]%=10;
            a[j+1]++;
        }
    }
    c[0]=j;
    if(c[j+1]>0) c[0]++;
    for(int i=c[0]; i>=1; i--) {
        if(x==false&&c[i]==0) continue;
        x=true;
        cout<<c[i];
    }
    printf("\n");
    return 0;
}

本蒟蒻的第一篇题解

结构体+选排

#include<bits/stdc++.h>
using namespace std;
struct eee
{
    char xibu,ganbu;
    string name;
    int pjcj,pycj,lws;
    long long RMB=0;
};
eee a[101];
int n;
long long sum=0;
int main()
{
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        cin>>a[i].name;
        cin>>a[i].pjcj>>a[i].pycj;
        cin>>a[i].ganbu>>a[i].xibu;
        cin>>a[i].lws;
        if(a[i].lws>=1&&a[i].pjcj>=81) a[i].RMB+=8000;
        if(a[i].pjcj>85&&a[i].pycj>80) a[i].RMB+=4000;
        if(a[i].pjcj>90) a[i].RMB+=2000;
        if(a[i].pjcj>85&&a[i].xibu=='Y') a[i].RMB+=1000;
        if(a[i].ganbu=='Y'&&a[i].pycj>80) a[i].RMB+=850;
        sum+=a[i].RMB;
    }
    for(int i=1;i<=n-1;i++)
        for(int j=i+1;j<=n;j++)
            if(a[i].RMB<a[j].RMB)
            {
                swap(a[i].ganbu,a[j].ganbu);
                swap(a[i].lws,a[j].lws);
                swap(a[i].name,a[j].name);
                swap(a[i].pjcj,a[j].pjcj);
                swap(a[i].pycj,a[j].pycj);
                swap(a[i].RMB,a[j].RMB);
                swap(a[i].xibu,a[j].xibu);
            }
    cout<<a[1].name<<endl;
    cout<<a[1].RMB<<endl;
    cout<<sum<<endl;
    return 0;
}

dalao们求一发模拟退火的题解OuO

本题链接：https://vijos.org/p/1000
JieKe08的第1篇题解！

for beginners，特设此题，^_^

#include<bits/stdc++.h>//多美妙的万能头
using namespace std;
int main()
{
    int a,b;//定义变量
    cin>>a>>b;//输入
    cout<<a+b;//输出，最后的换行可有可无
    return 0;//题目给的代码没有这个，但考试最好加上 
}

感谢您花费1分钟阅读鄙人的题解！

快读你们懂得吧
cpp
#include<bits/stdc++.h>
using namespace std;
int a,b;
inline int read(){
int ret=0,f=1;char ch=getchar();
while(!isdigit(ch)) f=(ch=='-'?-f:f),ch=getchar();
while(isdigit(ch)) ret=ret*10+ch-'0',ch=getchar();
return ret*f;
}
int main(){
a=read(),b=read();
printf("%d\n",a+b);
return 0;
}


//最简做法
#include<cstdio>
int main()
{
    int a,b;
    scanf("%d%d",&a,&b);
    printf("%d",a+b);
    return 0;
}

#include <bits/stdc++.h>
using namespace std;
int main(){
    int a,b;
    cin>>a>>b;
    cout<<a+b;
    return 0;
}

Python3
cpp
a,b = map(int,input().split())
print(a + b)