#include<bits/stdc++.h>
int main(int a,int b,int s)
{
if (!s)
{
printf("%d",b==0?a:main(a^b,(a&b)<<1,0));
exit(0);
}
scanf("%d%d",&a,&b);
main(a,b,0);
}
23333333333333333

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

#include<stdio.h>
int main( )
{
int a,b,c;
scanf("%d %d",&a,&b);
printf("%d",a+b);
return 0;
}

#include<stdio.h>
int main( )
{
int a,b,c;
scanf("%d %d",&a,&b);
printf("%d",a+b);
return 0;
}

#include<stdio.h>
int main()
{
int a,b;
scanf("%d%d",&a,&b);
printf("%d",a/b);
return 0;
}

#include <iostream>
#define a using
#define b namespace
#define c std
#define d ;
#define e int
#define f main
#define g (
#define h )
#define i {
#define j }
#define k cin
#define l >>
#define m return
#define n ,
#define o cout
#define p <<
#define q endl
#define r +
#define s 0

a b c d

e f g h
i
    e aaa n bbb d
    k l aaa l bbb d
    o p aaa r bbb p q d
    m s d
j

加强版

#include <iostream>
#define a using
#define b namespace
#define c std
#define d ;
#define e int
#define f main
#define g (
#define h )
#define i {
#define j }
#define k cin
#define l >>
#define m return
#define n ,
#define o cout
#define p <<
#define q endl
#define r +
#define s 0

a b c d

e f g h
{
    e aaa n bbb d
    k l aaa l bbb d
    o p aaa r bbb p q d
    m s d
}

我居然笑了

#include<cstdio>
int plus(int a,int b)
{
return b==0?a:plus(a^b,(a&b)<<1);
}
int main()
{
int a,b;scanf("%d%d",&a,&b);
printf("%d",plus(a,b));
}
位运算

#include<cstdio>
int plus(int a,int b)
{
if (b==0) return a;
plus(a^b,(a&b)<<1);
}
int main()
{
int a,b;scanf("%d%d",&a,&b);
printf("%d",plus(a,b));
}
23333

我才是最Diao的

program 呵呵;
var a,b:longint;
begin
  randomize;
  readln(a,b);
  a:=random(60000);
  writeln(a);
end.

var a,b:integer;
begin
read(a,b);
write(a+b);
end.

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
using namespace std;
int a[1001][1001];
int f[1001][1001];
int max(int a, int b){
return a > b ? a : b;
}
int main(){
memset(a, 0, sizeof(a));
memset(f, 0, sizeof(f));
int i,j,k,n,m;
scanf("%d%d", &n, &m);
for(i = 1; i <= n; i++)
for(j = 1; j <= m; j++) scanf("%d", &a[i][j]);
for(i = 1; i <= n; i++){
for(j = 1; j <= m; j++){
int tmp = max(f[i - 1][j] + a[i][j], f[i][j - 1] + a[i][j]);
f[i][j] = (f[i][j], tmp);
}
}
printf("%d\n", f[n][m]);
return 0;
}

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
using namespace std;
int a[100001];
int f[100001];
int max(int a, int b){
return a > b ? a : b;
}
int pre[100001];
int main(){
memset(pre, 0, sizeof(pre));
int i,j,k,n,m;
int ans = 0;
scanf("%d", &n);
for(i = 1; i <= n; i++) scanf("%d", &a[i]);
f[0] = 0; f[1] = 1; pre[1] = 1;
for(i = 2; i <= n; i++){
int MAXN = 0;
ans = 0;
for(j = 1; j <= i; j++){
if(a[j] <= a[i] && f[j] > MAXN){MAXN = f[j]; ans = j;}
}
f[i] = max(MAXN + 1, 1);
if(ans == 0) pre[i] = i;
else pre[i] = ans;
}
printf("%d\n", f[n]);
int MAXN = 0;
for(i = n; i >= 1; i--){
if(f[i] > MAXN){
MAXN = f[i];
ans = i;
}
}
printf("%d ", a[ans]);
for(i = ans; i != pre[i]; i = pre[i]){
printf("%d ", a[pre[i]]);
}
//for(i = 1; i <= n; i++) cout<<i<<": "<<pre[i]<<endl;
printf("\n");
return 0;
}
/*
6
5 3 4 8 6 7
*/

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define MAXN 1e16
using namespace std;
long long f[10001];
int a[10001];
int main(){
memset(f, 0, sizeof(f));
freopen("rand.out", "r", stdin);
freopen("硬币.out", "w", stdout);
int i,j,k,n,m,x,y,z;
scanf("%d", &m);
for(i = 1; i <=m; i++) scanf("%d", &a[i]);
scanf("%d", &n);
f[0] = 0;
for(i = 1; i <= n; i++){
long long minn = MAXN;
for(j = 1; j <= m && a[j] <= i; j++){
//cout<<i<<": "<<f[i - a[j]]<<endl;
if(f[i - a[j]] < minn && a[j] <= i){
minn = f[i - a[j]] + 1;
//cout<<i<<": "<<1<<endl;
}
}
f[i] = minn;
//else if(flag == 0)
//cout<<i<<": "<<f[i]<<endl;
}
if(f[n] == MAXN) printf("Impossible\n");
else printf("%lld\n", f[n]);
return 0;
}

#include<stdio.h>
int main()
{
int a,b;
scanf("%d%d",&a,&b);
printf("%d\n",a+b);
}
你们为什么这么diao

#include <stdio.h>
int main()
{
    int a, b;
    scanf("%d%d", &a, &b);
    printf("%d\n", a + b);
    return 0;
}

A+B问题，一道历史悠久的问题，下面就给出该题的多种解法。

解法一：首先用cin/scanf读入两个int类型数a和b，再用cout/printf输出a+b的结果。核心代码：cin>>a>>b; cout<<a+b<<endl;

解法二：为了防止计算a+b时占用过多内存（其实也没多少），可使用C++的提供的内联汇编来完成本题。核心代码：

__asm
(
"mov %1,%%eax \n\t"
"mov %2,%%ebx \n\t"
"add %%eax,%%ebx \n\t"
"mov %%ebx,%0"
:"=m"(b)
:"m"(a),"m"(b)
);
最后输出b即可。（事实上，这就是不让编译器替我们把高级语言翻译为汇编语言，直接就替它做了）

解法三：为防止出题人故意出大数据恶心我们，可采用高精度加法完成此题。核心代码：

char s1[101] , s2[101] , l1 , l2 , l3;
int a[101]={0} , b[101]={0} , c[102]={0};
cin >> s1 >> s2;
l1 = strlen(s1);
l2 = strlen(s2);
for (int i = 0 ; i < l1 ; i++) a[l1 - i] = s1[i] - '0';
for (int i = 0 ; i < l2 ; i++) b[l2 - i] = s2[i] - '0';
l3 = max(l1 , l2);
for (int i = 1 ; i <= l3 ; i++)
{
c[i] += a[i] + b[i];
c[i + 1] = c[i] / 10;
c[i] %= 10;
}
if (c[l3 + 1] == 1) l3++;
for (int i = l3 ; i >= 1 ; i--) cout << c[i]; cout<<endl;

解法四：（从此以后均为扯淡）

设f[i][j]表示i+j的值。可得状态转移方程为f[i][j] = max(f[i-1][j]+1 , f[i][j-1]+1)。由于任何数加0数值不变（公理），所以边界条件为f[0][i]=f[i][0]=i。最后输出f[a][b]即可。核心代码：

for (int i = 1 ; i <= a ; i++) f[i][0] = i;
for (int i = 1 ; i <= b ; i++) f[0][i] = i;
for (int i = 1 ; i <= a ; i++)
for (int j = 1 ; j <= b ; j++)
f[i][j] = max(f[i - 1][j] + 1 , f[i][j - 1] + 1);

解法五：分析解法四时间复杂度知若时间为O(a*b)。若a*b过大则程序必定超时。观察代码容易发现，根据分配律（公理）得：

f[i][j-1]+1=(i)+(j-1)+1=(i)+(j)-1+1=(i-1)+(j)+1=f[i-1][j]+1

根据上式我们可以立即将时间和空间复杂度都降低一维。核心代码：

f[0] = 0;
for (int i = 1 ; i <= max(a , b) ; i++)
f[i] = f[i - 1] + 1;

最后输出f[a]+f[b]。可以看到，时间复杂度为O(max(a,b))，为线性，可以通过此题。

解法六：

首先orz给出本解法的神犇

设图G=(V, E)为一流网络，包含2个源点S1、S2和1个汇点T。构造该图使得：

从S1、S2分别发出a、b条边，每条边的容量为1，这些边应当连接到max(a, b)个顶点上，且没有两条边拥有相同的两个端点。从这max(a, b)个顶点分别连一条容量为+∞的边到汇点T。然后在这个流网络上面跑Dinic算法、Ford-Fulkerson算法之类的东西，可以练练写网络流的手感。

Java作死版
```java
import java.math.BigDecimal;
import java.util.*;

public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
BigDecimal a = BigDecimal.valueOf(scanner.nextLong());
BigDecimal b = BigDecimal.valueOf(scanner.nextLong());
for(;b.longValue()>0;b=b.subtract(BigDecimal.ONE)){
a = a.add(BigDecimal.ONE);
}
System.out.println(a.toString());
scanner.close();
}
}
```

var a,b:longint;
begin
    readln(a,b);
    writeln(a+b);
end.

import java.io.*;
import java.util.*;

public class Main
{
public static void main (String args[]) throws Exception
{
BufferedReader stdin =
new BufferedReader(
new InputStreamReader(System.in));

String line = stdin.readLine();
StringTokenizer st = new StringTokenizer(line);
int a = Integer.parseInt(st.nextToken());
int b = Integer.parseInt(st.nextToken());
System.out.println(a+b);
}
}