题目链接 Cloud of Hashtags 题目还是比较简单的,直接贪心,但是因为我有两个细节没注意,所以FST了: 1.用了cin读入,但是没有加 std::ios::sync_with_stdio(false); 这条语句: 2.开了太多string. 也算是经验教训吧. #include <bits/stdc++.h> using namespace std; #define rep(i, a, b) for(int i(a); i <= (b); ++i) #define de…

http://codeforces.com/problemset/problem/777/D 题意:给出n道字符串,删除最少的字符使得s[i] <= s[i+1]. 思路:感觉比C水好多啊,大概是题目比较难看懂吧.直接从后面往前扫,用后面的答案更新前面的答案.考虑如果后面的字符串比前面的大,那么直接保存当前的字符串,否则暴力扫一遍,前面的字符串大于后面的字符串的那一位直接跳出. #include <bits/stdc++.h> using namespace std; ]; ]; int…

B. The Time 题目连接: http://www.codeforces.com/contest/622/problem/B Description You are given the current time in 24-hour format hh:mm. Find and print the time after a minutes. Note that you should find only the time after a minutes, see the examples t…

A. Infinite Sequence 题目连接: http://www.codeforces.com/contest/622/problem/A Description Consider the infinite sequence of integers: 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5.... The sequence is built in the following way: at first the number 1 is wr…

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…

Description Farmer John最近发明了一个游戏,来考验自命不凡的贝茜.游戏开始的时 候,FJ会给贝茜一块画着N (2 <= N <= 200)个不重合的点的木板,其中第i个点 的横.纵坐标分别为X_i和Y_i (-1,000 <= X_i <=1,000: -1,000 <= Y_i <= 1,000). 贝茜可以选两个点画一条过它们的直线,当且仅当平面上不存在与画出直线 平行的直线.游戏结束时贝茜的得分,就是她画出的直线的总条数.为了在游戏 中胜出,…

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…

题目链接: A. Vanya and Fence time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Vanya and his friends are walking along the fence of height h and they do not want the guard to notice them. In orde…

题意:给定 n 条边,判断是不是树. 析:水题,判断是不是树,首先是有没有环,这个可以用并查集来判断,然后就是边数等于顶点数减1. 代码如下: #include <bits/stdc++.h> using namespace std; const int maxn =1000 + 5; int p[maxn]; int Find(int x){ return x == p[x] ? x : p[x] = Find(p[x]); } int main(){ int n, m, x, y; cin…

https://codeforc.es/contest/1194/problem/B 好像也没什么思维,就是一个水题,不过蛮有趣的.意思是找缺黑色最少的行列十字.用O(n)的空间预处理掉一维,然后用O(n)的时间根据另一维计算出答案. #include<bits/stdc++.h> using namespace std; typedef long long ll; int n, m; string g[50005]; int rq[50005]; int main() { #ifdef Yi…