题目传送门 /* 很简单的水题,晚上累了,刷刷水题开心一下:) */ #include <bits/stdc++.h> using namespace std; ][] = {"zero", "one", "two", "three", "four", "five", "six", "seven", "eight",…

A. Tavas and Nafas Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/535/problem/A Description Today Tavas got his test result as an integer score and he wants to share it with his girlfriend, Nafas. His phone operating syste…

题目传送门 /* 题意:给定一个数列,求最大的r使得[l,r]的数字能在t次全变为0,每一次可以在m的长度内减1 二分搜索:搜索r,求出sum <= t * m的最大的r 详细解释:http://blog.csdn.net/libin56842/article/details/45082747 */ #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> us…

题目传送门 /* DFS:按照长度来DFS,最后排序 */ #include <cstdio> #include <algorithm> #include <cstring> #include <iostream> #include <cmath> #include <vector> using namespace std; ; const int INF = 0x3f3f3f3f; ] = {, }; int a[MAXN]; in…

题目传送门 /* 题意:一个字符串分割成k段,每段开头字母不相同 水题:记录每个字母出现的次数,每一次分割把首字母的次数降为0,最后一段直接全部输出 */ #include <cstdio> #include <iostream> #include <cstring> #include <string> #include <algorithm> using namespace std; ; const int INF = 0x3f3f3f3f;…

题目传送门 /* 水题:ans = (1+2+3+...+n) * k - n,开long long */ #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> using namespace std; typedef long long ll; int main(void) //Codeforces Round #304 (Div. 2) A. Soldier…

题目传送门 /* 题意:5种情况对应对应第i或j辆车翻了没 水题:其实就看对角线的上半边就可以了,vis判断,可惜WA了一次 3: if both cars turned over during the collision. 是指i,j两辆车,而不是全部 */ #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> #include <iostream>…

题目传送门 /* 水题:vector容器实现插入操作,暴力进行判断是否为回文串 */ #include <cstdio> #include <iostream> #include <algorithm> #include <cstring> #include <string> #include <vector> using namespace std; ; const int INF = 0x3f3f3f3f; vector<c…

题目传送门 /* 构造水题:对于0的多个位数的NO,对于位数太大的在后面补0,在9×k的范围内的平均的原则 */ #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> using namespace std; ; const int INF = 0x3f3f3f3f; int a[MAXN]; int main(void) //Codeforces Round #…

题目传送门 /* 水题:遍历一边先找AB,再BA,再遍历一边先找BA,再AB,两种情况满足一种就YES */ #include <cstdio> #include <iostream> #include <cstring> #include <cmath> #include <algorithm> #include <vector> #include <map> #include <queue> #includ…