41Acode 好像只要前一个字符串存在下一个字符串的头单词就YES: #include <bits/stdc++.h> using namespace std; typedef __int64 LL; const int N=1e2+10; int main() { int len,i,j; char s1[N],s2[N]; scanf("%s%s",s1,s2); len=strlen(s1); for(i=0,j=len-1;i<len;i++,j--) {…

题目背景 做正经题是不可能做正经题的,这辈子都不可能做正经题的,毒瘤题又不会做毒瘤题,就是水题这种东西,才维持了蒟蒻的信心: 题目描述 这里有N+1 道水题,编号分别为0 ~N+1 ,每道水题都有它自己水的程度,具体为对应编号的每位上的数字的乘积,现在为了能够更好的刷水题,我们需要统计一点小东西,统计给出L,R,A,B四个值,具体统计的内容为编号在LL 到RR 这段区间中水的程度在A 到B 之间的水题的编号和,现在请你编写一个程序来完成这个统计并A掉这道水题,因为结果可能很大,请输出在对P 取模…

L. Ministry of Truth Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100187/problem/K Description Andrey works in the Ministry of Truth. His work is changing articles in newspapers and magazines so that they praise the Party an…

水题系列 时间限制:1000 ms  |  内存限制:65535 KB 难度:2   描述     给你一个有向图,每条边都有一定的权值,现在让你从图中的任意一点出发,每次走的边的权值必须必上一次的权值大的情况下,问你最多能走几条边?   输入 首先一个n和m,分别表示点的数目和边的数目接下来m行,每行三个值x,y,val,表示x到y有路,其权值为val.(1<n,m,val<10^5,0<x,y<=n) 输出 输出最多有的边的数目 样例输入 3 3 1 2 1 2 3 1 3 1…

B. Archer Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/312/problem/B Description SmallR is an archer. SmallR is taking a match of archer with Zanoes. They try to shoot in the target in turns, and SmallR shoots first. The…

A. Fashion in Berland 题目连接: http://www.codeforces.com/contest/691/problem/A Description According to rules of the Berland fashion, a jacket should be fastened by all the buttons except only one, but not necessarily it should be the last one. Also if…

A. Opponents 题目连接: http://www.codeforces.com/contest/688/problem/A Description Arya has n opponents in the school. Each day he will fight with all opponents who are present this day. His opponents have some fighting plan that guarantees they will win…

后天考试,今天做题,我真佩服自己... 这次又只A俩水题... orz各路神犇... 话说这次模拟题挺多... 半个多小时把前面俩水题做完,然后卡C,和往常一样,题目看懂做不出来... A: 算是模拟吧,反正看懂题目后很可耻的生硬水果,没被hack我觉得非常开心... 由于要求两个人至少一个没跳过,画图后发现很偷懒的方法. 虽然跟样例不同,但过了... #include <cstdio> int main() { int n, m; scanf("%d%d", &n…

对于这道水题本人觉得应该应用贪心算法来解这道题: 下面就贴出本人的代码吧: #include<cstdio> #include<iostream> using namespace std; ],b[]; int main(void) { int n; ; ,sum2 = ; ;i<=;++i){ scanf("%d",&a[i]); sum1 += a[i]; } ;i<=;++i){ scanf("%d",&b[…

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…