A. Divisibility Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/597/problem/A Description Find the number of k-divisible numbers on the segment [a, b]. In other words you need to find the number of such integer values x tha…

A. Towers 题目连接: http://www.codeforces.com/contest/37/problem/A Description Little Vasya has received a young builder's kit. The kit consists of several wooden bars, the lengths of all of them are known. The bars can be put one on the top of the other…

A. Winner 题目连接: http://www.codeforces.com/contest/2/problem/A Description The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the max…

C. Subsequences Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/597/problem/C Description For the given sequence with n different elements find the number of increasing subsequences with k + 1 elements. It is guaranteed tha…

B. Restaurant Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/597/problem/B Description A restaurant received n orders for the rental. Each rental order reserve the restaurant for a continuous period of time, the i-th order…

C. Subsequences     For the given sequence with n different elements find the number of increasing subsequences with k + 1 elements. It is guaranteed that the answer is not greater than 8·1018. Input   First line contain two integer values n and k (1…

题目链接:http://codeforces.com/problemset/problem/328/B 水题~ #include <cstdio> #include <cstdlib> #include <cstring> ], a[]; ], c[]; int main(void) { //freopen("in.txt", "r", stdin); scanf("%s%s", n, a); int len1…

Codeforces Beta Round #12 (Div 2 Only) http://codeforces.com/contest/12 A 水题 #include<bits/stdc++.h> using namespace std; #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 #define sqr(x) ((x)*(x)) #define maxn 1000010 typedef long lo…

A.Beru-taxi 水题:有一个人站在(sx,sy)的位置,有n辆出租车,正向这个人匀速赶来,每个出租车的位置是(xi, yi) 速度是 Vi;求人最少需要等的时间: 单间循环即可: #include<iostream> #include<algorithm> #include<string.h> #include<stdio.h> #include<math.h> #include<vector> using namespace…

水题,只要遍历一遍,不够平均数的,从后面的借,比平均数多的,把多余的数添加到后面即可,注意数据范围 #include <iostream> #include <vector> #include <cmath> using namespace std; int main(){ int n; cin >> n; vector<long long> a(n); ; ; i < n; ++ i){ cin >>a[i]; sum +=a…