二分求解即可

#include<bits/stdc++.h>
#define N 100010
using namespace std;
int a,b;
int s(int a,int b) {
	int l=-1e9,r=1e9,mid;
	while(l<r) {
    	mid=(l+r)/2;
    	if(mid==a+b)
    	    return mid;
    	else if(mid>a+b)
    	    r=mid;
    	else
    	    l=mid;
    }
}
int main() {
	cin>>a>>b;
	cout<<s(a,b);
	return 0;
}

用线段树+标记永久化即可

#include <cstdio>
#define mid L + (R-L >> 1)
const int maxn = 1e5+5;

int n, a[maxn], m;
int sl, sr, add;

struct segtree{
	int sum[maxn<<2], tag[maxn<<2];
	inline int lc(int o){return o<<1;}
	inline int rc(int o){return o<<1|1;}
	void build(int o, int L, int R){
		if(L == R){sum[o] = a[L];return;}
		int M = mid;
		build(lc(o), L, M);
		build(rc(o), M+1, R);
		sum[o] = sum[lc(o)] + sum[rc(o)];
	}
	void maintain(int o, int L, int R){
		if(R>L){
			sum[o] = sum[lc(o)] + sum[rc(o)];
			sum[o] += tag[o] * (R-L+1);
		} else {
			sum[o] += tag[o];
			tag[o] = 0;
		}
	}
	void updata(int o, int L, int R){
		if(sl <= L && R <= sr)tag[o] += add;
		else{
			int M = mid;
			if(sl <= M)updata(lc(o), L, M);
			if(sr > M)updata(rc(o), M+1, R);
		}
		maintain(o, L, R);
	}
	int query(int o, int L, int R, int tags){
		if(sl <= L && R <= sr)return sum[o] + tags * (R-L+1);
		else {
			int M = mid, res = 0;
			if(sl <= M)res += query(lc(o), L, M, tags+tag[o]);
			if(sr > M)res += query(rc(o), M+1, R, tags+tag[o]);
			return res;
		}
	}
} sol;

signed main(){
	n = 1; 
	int a, b; 
	scanf("%d%d", &a, &b);
	sol.build(1, 1, n);
	add=a; sl=1; sr=1;
	sol.updata(1, 1, 1);
	add=b;
	sol.updata(1, 1, 1);
	printf("%d\n", sol.query(1, 1, n, 0));
	return 0;
}

当然，树状数组也可以，并且更快

#include <iostream>
using namespace std;
const int n = 1;
int a, b;
int c[500005];
inline int lowbit(int x){
	return x & (-x);
}
inline int sum(int x){
	int ans=0;
	for(int i=x;i>0;i-=lowbit(i))
	ans+=c[i];
	return ans;
}
void add(int x,int y){
	for(int i=x;i<=n;i+=lowbit(i))
	c[i]+=y;
}
int main(){
	cin>>a>>b;
	add(1, a); add(1, b);
	printf("%d\n", sum(1));
	return 0;
}

高精度水一发......

#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
string a,b;
int x[1005],y[1005],ans[1005];
int main()
{
	cin>>a>>b;
	int la=a.size(),lb=b.size();
	reverse(a.begin(),a.end());
	reverse(b.begin(),b.end());
	int n,i,j;
	for(i=0;i<la;i++)    x[i]=a[i]-'0';
	for(i=0;i<lb;i++)    y[i]=b[i]-'0';
	int temp,jin=0;
	for(i=0;i<1000;i++)
	{
		temp=x[i]+y[i]+jin;
		if(temp>=10)
		{
			jin=1;
			ans[i]=temp-10;
		}
		if(temp<10)
		{
			jin=0;
			ans[i]=temp;
		}
	}
	for(i=1000;i>=0;i--)
		if(ans[i]!=0)
			break;		
	for(;i>=0;i--)
	{
		cout<<ans[i];
	}
    return 0;
}

输出A+B非常简单
#include <bits/stdc++.h> using namespace std; int a,b; cin>>a>>b; cout<<a+b<<endl;//输出； return 0;//华丽结束

这题挺简单的，是个入门题

不说了，直接上code：

#include<bits/stdc++.h>

using namespace std;

int a,b;//定义

int main() {

cin>>a>>b;//输入
cout<<a+b;//输出

}

#include #define mid L + (R-L >> 1) const int maxn = 1e5+5;