48 条题解
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lintaicheng LV 5 @ 2026-4-15 13:26:15
#include<stdio.h> int main(){ int a,b; scanf("%d%d",&a,&b); printf("%d",a+b); } -
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a=int(input()) b=int(input()) print(a+b) -
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jianbaitian LV 7 @ 2026-3-22 17:42:07
#include <bits/stdc++.h> using namespace std; int main(){ int a,b,s; cin>>a>>b; s=a+b; cout<<s; return 0; } -
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#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; -
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#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; } -
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#include<bits/stdc++.h> using namespace std; const int N=1010;//1表示开头为1,2表示10的2次方 const int INT=0x3f3f3f3f;//INT+INT int范围内最大INT*INT ,long long; int n,m; void dfs(int n,int m){ int sum=0,ans=0; sum=n; ans=m; cout<<ans+1-1+1-1+1-1+sum+1-1+1-1+1-1; } int main( ) { cin>>n>>m; dfs(n,m); }
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#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; -
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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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chenzirui1 LV 6 @ 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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权威·忘本 @ 2025-5-11 9:37:50
权威
#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); //连边 Link connect(A,B); //断边 Cut cut(A,B); //再连边orz Link again connect(A,B); printf("%d\n",query(A,B)); return 0; }
zzq1234
LV 8
@ 2026-4-8 19:18:42
