A. Group of Students time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output At the beginning of the school year Berland State University starts two city school programming groups, for beginners an…

Codeforces Round #297 (Div. 2)A. Vitaliy and Pie Time Limit: 2 Sec  Memory Limit: 256 MBSubmit: xxx  Solved: 2xx 题目连接 http://codeforces.com/contest/525/problem/A Description After a hard day Vitaly got very hungry and he wants to eat his favorite pot…

A. Mahmoud and Longest Uncommon Subsequence time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output While Mahmoud and Ehab were practicing for IOI, they found a problem which name was Longest comm…

题目:http://codeforces.com/problemset/problem/356/A 题意:首先给你n,m,代表有n个人还有m次描述,下面m行,每行l,r,x,代表l到r这个区间都被x所击败了(l<=x<=r),被击败的人立马退出游戏让你最后输出每个人是被谁击败的,最后那个胜利者没被 人击败就输出0 思路:他的每次修改的是一个区间的被击败的人,他而且只会记录第一次那个被击败的人,用线段树堕落标记的话他会记录最后一次的,所以我们倒着来修改, 然后因为那个区间里面还包含了自己,在线段…

脑子又卡了...来一发set的,STL真心不熟. #include <stdio.h> #include <string.h> #include <iostream> #include <set> using namespace std; ]; set<int>se; set<int>::iterator it1,it,it2; int main() { int i,x,y,num,n,m; scanf("%d%d"…

#include <iostream> #include <vector> using namespace std; int main(){ ; cin >> m ; vector<,); ; i <= m ; ++ i) {cin >> c[i];sum+=c[i];} , secondPart = ,firstIndex = , secondIndex =; cin >> x >> y; ,secondIndex =…

数学so奇妙.. 这题肯定会有一个循环节 就是最小公倍数 对于公倍数内的相同的数的判断 就要借助最大公约数了 想想可以想明白 #include <iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<stdlib.h> #include<vector> using namespace std; #define N 1000100 #defin…

读错题意了..线段树延迟标记 白刷这么多线段树 #include <iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<stdlib.h> #include<vector> using namespace std; #define N 300010 ],la[N<<]; struct node { int l,r,d; }q[N…

A:超级大水题: 代码: #include<cstdio> #define maxn 105 using namespace std; int n,a[maxn],x,y,ans; int main() { scanf("%d",&n); ; i<=n; i++) { scanf("%d",&x); a[i]=a[i-]+x; ans+=x; } scanf("%d%d",&x,&y); int…

题目链接: B. Xenia and Hamming 题意: 要求找到复制后的两个字符串中不同样的字符 思路: 子问题: 在两串长度是最大公倍数的情况下, 求出一个串在还有一个串中反复字符的个数 CODE: #include <iostream> #include<stdio.h> #include<string> #include<string.h> using namespace std; #define M 1000006 int com[M][26]…