Wow! Such Doge! - HDU 4847 (水题)】的更多相关文章

http://acm.hdu.edu.cn/showproblem.php?pid=4847 Wow! Such Doge! Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 4847 Description Chen, Adrian (November 7, 2013). “Doge Is An Ac- tually Good Inter…
题目大意:题目描述了一大堆.....然而并没什么用,其实就是让求给的所有字符串里面有多少个"doge",不区分大小写.   代码如下: ====================================================================================================================== #include<stdio.h> #include<string.h> #include&…
就是求出现了多少次doge 不区分大小写  巧用字符串函数 isalpha 判断是否是字母 tolower 转换为小写字母 toupper 转换为大写字母 strncmp字符串比较函数  能限制比较的长度 #include <iostream> #include <cstdio> #include <sstream> #include <cstring> #include <map> #include <cctype> #includ…
N! Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others) Total Submission(s): 64256    Accepted Submission(s): 18286 Problem Description Given an integer N(0 ≤ N ≤ 10000), your task is to calculate N!   Input One N in…
E - 5 Time Limit:1500MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 5391 Description Tina Town is a friendly place. People there care about each other. Tina has a ball called zball. Zball is magic. It grows…
水题 /* * Author : ben */ #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <ctime> #include <iostream> #include <algorithm> #include <queue> #include <set> #include <m…
1040水题; These days, I am thinking about a question, how can I get a problem as easy as A+B? It is fairly difficulty to do such a thing. Of course, I got it after many waking nights.Give you some integers, your task is to sort these number ascending (…
题意:判断一些数里有最大因子的数 水题,省赛即将临近,高效的代码风格需要养成,为了简化代码,以后可能会更多的使用宏定义,但是通常也只是快速拿下第一道水题,涨自信.大部分的代码还是普通的形式,实际上能简化的部分也不太多 #include<iostream> #include<cstring> #include<cmath> #include<cstdio> using namespace std; #define for0n for(i=0;i<n;i+…
/* 对于只会弗洛伊德的我,迪杰斯特拉有点不是很理解,后来发现这主要用于单源最短路,稍稍明白了点,不过还是很菜,这里只是用了邻接矩阵 套模板,对于邻接表暂时还,,,没做题,后续再更新.现将这题贴上,应该是迪杰斯特拉最水的题没有之一.纯模板 找到距离起点最近的点,以此点为中间点进行更新,找到了在进行下一个点. */ 题目大意: 搬东西很累,想省力,给你几个点和点之间的距离:标准题型: #include<stdio.h> #include <iostream> #include<…
题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5162 题解:看了半天以为测试用例写错了.这题玩文字游戏.它问的是当前第i名是原数组中的第几个. #include<stdio.h> #include<iostream> #include<string.h> #include <stdlib.h> #include<math.h> #include<algorithm> #include…