快速切题 cf118A
这教导人们一定要看题,要看题,元音包含了‘y’,完毕,要看题啊
#include <cstring>
#include <cstdio>
#include <cctype>
using namespace std;
char buff[1001];
int main(){
scanf("%s",buff);
for(int i=0;buff[i];i++){
buff[i]=tolower(buff[i]);
if(buff[i]=='a'||buff[i]=='o'||buff[i]=='i'||buff[i]=='e'||buff[i]=='u'||buff[i]=='y'){continue;}
else {
printf(".%c",buff[i]);
}
} printf("\n");
return 0;
}
快速切题 cf118A的更多相关文章
- 快速切题sgu127. Telephone directory
127. Telephone directory time limit per test: 0.25 sec. memory limit per test: 4096 KB CIA has decid ...
- 快速切题sgu126. Boxes
126. Boxes time limit per test: 0.25 sec. memory limit per test: 4096 KB There are two boxes. There ...
- 快速切题 sgu123. The sum
123. The sum time limit per test: 0.25 sec. memory limit per test: 4096 KB The Fibonacci sequence of ...
- 快速切题 sgu120. Archipelago 计算几何
120. Archipelago time limit per test: 0.25 sec. memory limit per test: 4096 KB Archipelago Ber-Islan ...
- 快速切题 sgu119. Magic Pairs
119. Magic Pairs time limit per test: 0.5 sec. memory limit per test: 4096 KB “Prove that for any in ...
- 快速切题 sgu118. Digital Root 秦九韶公式
118. Digital Root time limit per test: 0.25 sec. memory limit per test: 4096 KB Let f(n) be a sum of ...
- 快速切题 sgu117. Counting 分解质因数
117. Counting time limit per test: 0.25 sec. memory limit per test: 4096 KB Find amount of numbers f ...
- 快速切题 sgu116. Index of super-prime bfs+树思想
116. Index of super-prime time limit per test: 0.25 sec. memory limit per test: 4096 KB Let P1, P2, ...
- 快速切题 sgu115. Calendar 模拟 难度:0
115. Calendar time limit per test: 0.25 sec. memory limit per test: 4096 KB First year of new millen ...
随机推荐
- SQLException: com.mchange.v2.c3p0.ComboPooledDataSource [ java.beans.IntrospectionException: java.lang.reflect.InvocationTargetException [numThreadsAwaitingCheckoutDefaultUser] ] has been closed()
问题:Could not get JDBC Connection; nested exception is java.sql.SQLException: com.mchange.v2.c3p0.Com ...
- 发现vi出现此错误~/.vim/bundle/YouCompleteMe/third_party/ycmd/ycm_core.so: undefined symbol: clang_getCompletionFixIt
答: 安装更高版本的clang库 sudo apt-get install clang-7 sudo update-alternatives --install /usr/bin/clang clan ...
- git源码阅读
https://github.com/git-for-windows/git/issues/1854 https://github.com/git-for-windows/git/pull/1902/ ...
- 04_Flume多节点load_balance实践
1.负载均衡场景 1)初始:上游Agent通过round_robin selector, 将event轮流发送给下游Collecotor1, Collector2 2)故障: 关闭Collector1 ...
- UVa 11100 旅行2007
https://vjudge.net/problem/UVA-11100 题意: 给定n个正整数,把它们划分成尽量少的严格递增序列,尽量均分. 思路: 因为必须严格递增,所以先统计每个数字出现的次数, ...
- codevs 1200 同余方程 逆元
题目描述 Description 求关于 x 同余方程 ax ≡ 1 (mod b)的最小正整数解. 输入描述 Input Description 输入只有一行,包含两个正整数 a, b,用 一个 空 ...
- 使用R的数据库查询
JS 很多方法可以用R查询数据.这篇文章展示了三种最常见的方法: 运用 DBI 使用dplyr语法 使用R note book 背景 最近的一些软件包改进可以更轻松地将数据库与R一起使用.下面的查询示 ...
- 让flask在出现语法错误时仍然自动重启
1问题描述: flask自带的reload只能在语法没毛病的情况下auto_relaod,但是如果有语法错误,进程就会报错退出. 这时修改完语法错误,还得在控制台按“↑”和“enter”重新执行一次p ...
- Codeforces 913C - Party Lemonade
913C - Party Lemonade 思路:对于第i个话费cost[i],取min(cost[i],2*cost[i-1]),从前往后更新,这样就可以保证第n个的话费的性价比最高,那么从最高位开 ...
- HDU 6130 Kolakoski
Kolakoski 思路: 从前往后扩展,前后构成映射关系. 代码: #include<bits/stdc++.h> using namespace std; #define ll lon ...