63 条题解
-
-4
huangzhihao LV 8 @ 2025-12-21 20:09:08
#include using namespace std; int main() { int a,b; cin>>a>>b; cout<<a+b;
}
-
-4
sng2025122
LV 6
@ 2025-12-13 19:56:30
#include using namespace std; int main() { int a , b;
cin>>a>>b; cout<<a+b; return 0;}
-
-4
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; } -
-5
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; } -
-6
http://ybt.ssoier.cn:8088/problem_show.php?pid=1510
#include <bits/stdc++.h> using namespace std; const int N = 1e5 + 10; const int INF = 0x3f3f3f3f; int T; int r[30];//r[i]表示第i个时间点需要工作的人数 int n; int num[30] , x;//num[i]表示第i个时间点有多少人开始工作 int L , R , ans; vector<pair<int,int> > vc[N]; int dis[N]; bool vis[N]; queue<int> q; void spfa(int mid)//最长路 { memset(dis, -INF , sizeof(dis)); memset(vis , 0 , sizeof(vis)); while(!q.empty()) q.pop(); dis[0] = 0; vis[0] = 1; q.push(0); while(!q.empty()) { int u = q.front(); q.pop(); if( u == 24 && dis[u] > mid) return; 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; } } } } } bool check(int mid)//一共mid人工作 { for(int i = 0; i <= 24; i++) { vc[i].clear(); } //隐藏不等式 //sum[i] 从1点到i点需要工作的人数 // sum[i] - sum[i - 1] >= 0 // sum[i - 1] - sum[i] >= -num[i] for(int i = 1; i <= 24; i++) { vc[i - 1].push_back({i , 0}); vc[i].push_back({i - 1 ,-num[i]}); } // 23 24 1 2 3 4 5 6 7 8 9 // sum[i] - sum[i - 8] >= r[i] for(int i = 8; i <= 24; i++) vc[i - 8].push_back({i , r[i]}); // sum[24] - sum[8] <= mid - r[i]; // sum[i] - sum[i + 16] >= r[i] - mid; for(int i = 1; i <= 8; i++) vc[i + 16].push_back({i , r[i] - mid}); // sum[24] - sum[0] <= mid vc[0].push_back({24,mid}); vc[24].push_back({0, -mid}); spfa(mid); return dis[24] == mid; } int main() { cin >> T; while( T-- ) { memset(num , 0 , sizeof(num)); for(int i = 1; i <= 24; i++) cin >> r[i]; cin >> n; //表示每个人开始工作的时间 for(int i = 1; i <= n; i++) { cin >> x; num[x + 1]++; } //二分答案 L = 0 , R = n , ans = -1; while( L <= R) { int mid = L + R >> 1; if(check(mid)) { ans = mid; R = mid - 1; } else L = mid + 1; } if(ans == -1) cout << "No Solution\n"; else cout << ans << endl; } return 0; } -
-6
sng2025102 LV 5 @ 2025-7-6 23:16:54
#include
using namespace std;
int main()
{
int a; int b; cin>>a>>b; cout<<a+b; return 0;}
-
-6
liangjingxu LV 7 @ 2024-11-10 20:23:47
#include<iostream> using namespace std; int main () { int a,b; cin >> a >> b; cout << a + b; }最简单的代码了，自己拿去用
-
-6
高精度加法
新人不会康林一个人 @杨时欢 俩提交都AC
阿米诺斯亲放心食用#include<bits/stdc++.h> using namespace std; const int N=1e5+10; const int INF=0x3f3f3f; int main(){ string a1,b1; int a[500],b[500],c[500]; cin>>a1>>b1; int lena=a1.size(); int lenb=b1.size(); for(int i=0;i<lena;i++){ a[lena-i]=a1[i]-'0'; } for(int i=0;i<lenb;i++){ b[lenb-i]=b1[i]-'0'; } int lenc=1,x=0; while(lenc<=lena || lenc<=lenb){ c[lenc]=a[lenc]+b[lenc]+x; x=c[lenc]/10; c[lenc]=c[lenc]%10; lenc++; } if(x>0){ c[lenc]=x; } else{ lenc--; } for(int i=lenc;i>=1;i--) cout<<c[i]; cout<<endl; return 0; }提示：此代码可提交题@@高精度加法*** -
-7
#A+B problem {
{ #include//头文件，使用的工具库为iostream using namespace std;//命名空间 int main()//主函数 //先问大家几个问题： //1.我们如何获取a与b？ //2.我们如何进行对a于b的运算? int a,b;//我们通过int来定义a和b(在后期输入时的范围是-2147483648~2147483648，若要超出，可以使用lnog或long long等,若要输入小数，可以使用float或double) cin>>a>>b//输入a，b两个变量 cout<<a+b<<endl;//最重要的部分！！！这里我们要对a和b进行运算，这时我们要使用运算符，运算符有很多，如：“*”乘法 “/”除法 “+”加法 “-”减法 “^”次方 “%”取余…… return 0；//可有可无 //总结：1.a与b通过int定义，且不同的类型变量有不同规定取值范围； //2.我们使用运算符进行两个变量的运算
-
-7
#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; }
huangzhihao
LV 8
@ 2025-12-21 20:11:01
