摘要： #include <iostream>#include<stack>#include<stdio.h>#include<string.h>#include<stdlib.h>#include<string>using namespace std;int N;string come;string leave;int order[50];int main(){ setbuf(stdout,NULL); stack<char>s; while(cin>>N) { cin>>come; cin& 阅读全文

摘要： #include<stdio.h>#include<cstdlib>#include <iostream>using namespace std;int GCD(int x,int y){ if(!x || !y)return x>y?x:y; for(int t;t=x%y;x=y,y=t); return y;}int LCM(int x,int y){ int gcd=GCD(x,y); return x/gcd*y;}int ans;int main(){ setbuf(stdout,NULL); int T,number,i,inter; s 阅读全文

摘要： #include<stdio.h>#include<string.h>#include <iostream>using namespace std;int N;int prime[12]={2,3,5,7,11,13,17,19,23,29,31,37};int ring[25];int visited[25];int search(int low ,int high,int x){ if(low>high)return 0; int mid=(low+high)>>1; if(x==prime[mid])return mid; else 阅读全文

摘要： 这道数学题用到了数论中的一个公式,叫做斯特林公式，lnN!=NlnN-N+0.5*ln(2*N*pi)//============================================================================// Name : HDU.cpp// Author : lonelycatcher// Version :// Copyright : Your copyright notice// Description : Hello World in C++, Ansi-style//================================ 阅读全文

摘要： 题意很简单，就是求一定范围内子矩阵的和，数据量很大，用暴力的话一定会超时，这道题用二维的树状数组做比较简单。。。#include<stdio.h>#include<string.h>#include <iostream>using namespace std;#define MAX 1030int tree[MAX][MAX];int S,ans;int lowbit(int x){return x&(-x);}void add(int x,int y,int num){ for(int i=x;i<=S;i+=lowbit(i)) { for 阅读全文