Online C Compiler: What's the Buzz All About? The convenience and flexibility provided by online C compilers have caught the attention of the programming realm. Here are some of the reasons for their increasing traction: Cost-Effective: Many online C compilers are available free of charge. This...
找个最大pig,然后所有比他小的其他种类生物一直加就好了 #include<bits/stdc++.h>usingnamespacestd;typedeflonglongll;constll mod=1e9+7;llksm(ll x,ll y){ ll ans=1;while(y) {if(y&1) ans=(ans%mod*x%mod)%mod; x=x%mod*(x%mod)%mod; y>>=1; }returnans%mod%mod; }voidfio(){ ...
l,r);if(l>mid)returnquery(u<<1|1,l,r);Nodeans,left=query(u<<1,l,r),right=query(u<<1|1,l,r);ans.l=left.l,ans.r=right.r;ans.sum=left.sum+right.sum;ans.lnum=left.lnum,ans.rnum=right.rnum;ans.
Participate in online C++ output contests to enhance your programming skills and compete with others. Find matches and improve your coding efficiency.
#include #include<cmath> #include<stack> #include<queue> #include<cstdio> #include<vector> #include<string> #include<cstring> #include<iostream> #include<algorithm> #define Endl "\n" using namespace std; typedef long long ll; typedef pair<int,int> pii; const int maxn = 2000007; const...
P.P.S. Don't blame me for not telling everyone during contest. I wasn't aware of it until the last day. Example: godbolt #include <cstdio> int main() { double val = 1337.1337221 ; const char *str_val = "1337.1337221"; double scanned; sscanf(str_val, "%lf", &scanned); if ...
The 2019 Asia Nanchang First Round Online Programming Contest C(cf原题,线段树维护矩阵) 题:https://nanti.jisuanke.com/t/41350 分析:先将字符串转置过来 状态转移,因为只有5个状态,所以 i 状态到 j 状态的最小代价就枚举【i】【k】->【k】【j】的最小值(0<=k<=4)...
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Java remains one of the most popular programming languages, offering numerous compelling advantages for both budding and seasoned developers. Its syntax is clear and, to some extent, influenced by C++, making the transition smoother for those familiar with C-like languages. Java is versatile, being...
The 2019 Asia Nanchang Online Programming Contest B. Fire-Fighting Hero 题目 题目不难,就是太难懂。 给一个无向图,有一个点 s,和 k 个点,问点 s 到其他点的最短路径的最大值,跟 k 个点到其他点最短路径的最大值。谁的小谁赢,不过比之前 s 得出的值要乘以 c 再比。输出赢家的值。