39 条题解
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0
chenzirui1 LV 5 @ 2026-3-16 21:13:55
#include <bits/stdc++.h> using namespace std;
int main(){ int a,b; cin>>a>>b; cout<<a+b; return 0; }
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0
#include<bits/stdc++.h> using namespace std; int main() { string a1,b1; int a[10000]={},b[10000]={},c[10000]={}; cin >> a1 >> b1; int lena=a1.size(); int lenb=b1.size(); for(int i=1;i<=lena;i++) { a[i]=a1[lena-i]-'0'; } for(int i=1;i<=lenb;i++) { b[i]=b1[lenb-i]-'0'; } int lenc=1; while(lenc<=lena||lenc<=lenb) { c[lenc]+=a[lenc]+b[lenc]; if(c[lenc]>9) { c[lenc]-=10; c[lenc+1]++; } lenc++; } if(c[lenc]==0) { lenc--; } for(int i=lenc;i>=1;i--) { cout << c[i]; } return 0; } -
-1
huangzhihao LV 9 @ 2025-12-21 20:11:01
#include using namespace std; int main() { int a,b; cin>>a>>b; cout<<a+b; }
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-1
#include using namespace std; int main() { int a,b; cin>>a>>b; cout<<a+b;
}
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-1
sng2025122
LV 7
@ 2025-12-13 19:56:30
#include using namespace std; int main() { int a , b;
cin>>a>>b; cout<<a+b; return 0;}
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-1
yangshihuan
LV 7
@ 2024-7-26 10:58:29
A+B Problem题解
新用户强烈建议阅读此帖
首先我们要理清思路
1.需要用到什么样的头文件？
2.用什么样的数据范围？
3.思路是什么？
首先题目中的数据范围是1≤a,b≤10^6, 而int 的范围是-2147483648-2147483647 正合题意，所以数据类型可以用int
话不多说，直接上代码
#include<iostream>//导入头文件，iostream里面是标准输入输出流（我说的什么？） using namespace std;//使用标准命名空间 int main(){//主函数，程序的入口 int a,b;//创建a,b两个整型变量 cin>>a>>b;//输入 a , b 两个变量 cout<<a+b; //输出a+b的内容 return 0; }本蒟蒻发的第一篇题解，请多多支持喵~~
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-2
#include<bits/stdc++.h> using namespace std; const int N=1e5+5; int n , m , a[N]; int u , v; vector<int> vc[N]; int dfn[N] , low[N] , cnt; bool vis[N]; stack<int> st; int belong_cnt; //强连通分量个数 vector<int> belong[N];//强连通分量 int color[N];//color[i]表示第i头牛所处的强连通分量 vector<int> newvc[N]; int in[N];//入度 int ans[N]; void tarjan(int u) { dfn[u] = low[u] = ++cnt; vis[u] = 1; st.push(u); for(int i = 0; i < vc[u].size(); i++) { int v = vc[u][i]; if(!dfn[v]) { tarjan(v); low[u] = min(low[u] , low[v]); } else if(vis[v]) { low[u] = min(low[u] , low[v]); } } //强连通分量到头 if(dfn[u] == low[u]) { while(!st.empty()) { v = st.top(); st.pop(); vis[v] = 0; color[v] = u;//当前牛所处的强连通分量 if(u == v) break; a[u] += a[v]; } } } void tupo() { queue<int> q; for(int i = 1; i <= n; i++) { if(color[i] == i && !in[i]) { q.push(i); ans[i] = a[i]; } } while(!q.empty()) { int top = q.front(); q.pop(); for(int i = 0; i < newvc[top].size(); i++) { v = newvc[top][i]; ans[v] = max(ans[v] , ans[top] + a[v]); in[v]--; if(!in[v]) q.push(v); } } int maxx = 0; for(int i = 1; i <= n; i++) maxx = max(maxx , ans[i]); cout << maxx; } int main(){ cin >> n >> m; for(int i = 1; i <= n; i++) cin >> a[i]; while( m-- ) { cin >> u >> v; vc[u].push_back(v); } for(int i = 1; i <= n; i++) if(!dfn[i]) tarjan(i); //重新建边 for(int i = 1; i <= n; i++) { for(int j = 0; j < vc[i].size(); j++) { v = vc[i][j]; if(color[i] != color[v]) { newvc[color[i]].push_back(color[v]); in[color[v]]++; } } } tupo(); return 0; -
-2
#include<bits/stdc++.h> using namespace std; const int N=1e5+5; int n , m , a[N]; int u , v; vector<int> vc[N]; int dfn[N] , low[N] , cnt; bool vis[N]; stack<int> st; int belong_cnt; //强连通分量个数 vector<int> belong[N];//强连通分量 int color[N];//color[i]表示第i头牛所处的强连通分量 vector<int> newvc[N]; int in[N];//入度 int ans[N]; void tarjan(int u) { dfn[u] = low[u] = ++cnt; vis[u] = 1; st.push(u); for(int i = 0; i < vc[u].size(); i++) { int v = vc[u][i]; if(!dfn[v]) { tarjan(v); low[u] = min(low[u] , low[v]); } else if(vis[v]) { low[u] = min(low[u] , low[v]); } } //强连通分量到头 if(dfn[u] == low[u]) { while(!st.empty()) { v = st.top(); st.pop(); vis[v] = 0; color[v] = u;//当前牛所处的强连通分量 if(u == v) break; a[u] += a[v]; } } } void tupo() { queue<int> q; for(int i = 1; i <= n; i++) { if(color[i] == i && !in[i]) { q.push(i); ans[i] = a[i]; } } while(!q.empty()) { int top = q.front(); q.pop(); for(int i = 0; i < newvc[top].size(); i++) { v = newvc[top][i]; ans[v] = max(ans[v] , ans[top] + a[v]); in[v]--; if(!in[v]) q.push(v); } } int maxx = 0; for(int i = 1; i <= n; i++) maxx = max(maxx , ans[i]); cout << maxx; } int main(){ cin >> n >> m; for(int i = 1; i <= n; i++) cin >> a[i]; while( m-- ) { cin >> u >> v; vc[u].push_back(v); } for(int i = 1; i <= n; i++) if(!dfn[i]) tarjan(i); //重新建边 for(int i = 1; i <= n; i++) { for(int j = 0; j < vc[i].size(); j++) { v = vc[i][j]; if(color[i] != color[v]) { newvc[color[i]].push_back(color[v]); in[color[v]]++; } } } tupo(); return 0; -
-2
mengqingyu
LV 10
@ 2024-11-16 16:21:16
#include<iostream> using namespace std; int main(){ int a,b,c; cin>>a>>b; c=a+b; cout<<c; } -
-3
http://ybt.ssoier.cn:8088/problem_show.php?pid=1509
#include <bits/stdc++.h> using namespace std; const int N = 5e4 + 10; const int INF = 0x3f3f3f3f; int n; int u , v , w , maxx; vector<pair<int,int> > vc[N]; int dis[N]; bool vis[N]; void spfa()//求最长路！！！ { memset(dis, -INF, sizeof(dis)); dis[0] = 0; vis[0] = 1;//表示当前点是否在队列中 queue<int> q; q.push(0); while(!q.empty()) { int u = q.front(); q.pop(); vis[u] = 0; for(int i = 0; i < vc[u].size(); i++) { int v = vc[u][i].first , w = vc[u][i].second; if(dis[v] < dis[u] + w) { dis[v] = dis[u] +w; if(!vis[v]) { q.push(v); vis[v] = 1; } } } } } int main() { cin >> n; for(int i = 1; i <= n; i++) { cin >> u >> v >> w; u++ , v++;//整体右移 //sum[v] - sum[u - 1] >= w vc[u - 1].push_back({v , w}); maxx = max(maxx , v); } //隐藏不等式 sum[i] - sum[i - 1] >= 0 sum[i - 1] - sum[i] >= -1 for(int i = 1; i <= maxx; i++) { vc[i - 1].push_back({i , 0}); vc[i].push_back({i - 1, -1}); } spfa(); cout << dis[maxx]; return 0; } -
-3
dingyizhen LV 8 @ 2025-11-22 9:41:59
#include<iostream> using amespace std; int vera(int x,int y){ return x+y; } int main(){ int a=1,b=2; cout<<vera(a,b); return 0; } -
-3
#include<iostream> using namespace std; int main(){ int a,b; cin>>a>>b; cout<<a+b; return 0; }
