Codeforces Testing Round #8 B. Sheldon and Ice Pieces 水题
题目链接:http://codeforces.com/problemset/problem/328/B
水题~
#include <cstdio>
#include <cstdlib>
#include <cstring>
char n[], a[];
int b[], c[];
int main(void) {
//freopen("in.txt", "r", stdin);
scanf("%s%s", n, a);
int len1 = strlen(n), len2 = strlen(a);
memset(b, , sizeof(b)); memset(c, , sizeof(c));
for (int i = ; i < len1; ++i) b[n[i]-'']++;
for (int i = ; i < len2; ++i) c[a[i]-'']++;
b[]+=b[], b[]=, c[]+=c[], c[]=, b[]+=b[], b[]=, c[]+=c[], c[]=;
int cnt=, Max=;
for (int i = ; i < ; ++i){
if (b[i]){
cnt = c[i]/b[i]; if (cnt < Max) Max = cnt;
}
}
printf("%d\n", Max);
return ;
}
=_=
Codeforces Testing Round #8 B. Sheldon and Ice Pieces 水题的更多相关文章
- Codeforces Beta Round #4 (Div. 2 Only) A. Watermelon 水题
A. Watermelon 题目连接: http://www.codeforces.com/contest/4/problem/A Description One hot summer day Pet ...
- Codeforces Beta Round #14 (Div. 2) C. Four Segments 水题
C. Four Segments 题目连接: http://codeforces.com/contest/14/problem/C Description Several months later A ...
- Codeforces Beta Round #14 (Div. 2) B. Young Photographer 水题
B. Young Photographer 题目连接: http://codeforces.com/contest/14/problem/B Description Among other thing ...
- Codeforces Beta Round #6 (Div. 2 Only) A. Triangle 水题
A. Triangle 题目连接: http://codeforces.com/contest/6/problem/A Description Johnny has a younger sister ...
- Codeforces 148D 一袋老鼠 Bag of mice | 概率DP 水题
除非特别忙,我接下来会尽可能翻译我做的每道CF题的题面! Codeforces 148D 一袋老鼠 Bag of mice | 概率DP 水题 题面 胡小兔和司公子都认为对方是垃圾. 为了决出谁才是垃 ...
- Codeforces Round #359 (Div. 2) A. Free Ice Cream 水题
A. Free Ice Cream 题目连接: http://www.codeforces.com/contest/686/problem/A Description After their adve ...
- Codeforces Testing Round #12 A. Divisibility 水题
A. Divisibility Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/597/probl ...
- Codeforces Round #334 (Div. 2) A. Uncowed Forces 水题
A. Uncowed Forces Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/604/pro ...
- Codeforces Round #115 B. Plane of Tanks: Pro 水题
B. Plane of Tanks: Pro Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/17 ...
随机推荐
- 服务器端与客户端TCP连接入门(一)
Java中使用Socket(即套接字)完成TCP程序的开发 服务器端使用ServerSocket接收客户端的连接请求,每一个客户端都使用一个Socket对象表示 在服务器端每次运行时都要使用accep ...
- Spark运行环境的安装
scala-2.9.3:一种编程语言,下载地址:http://www.scala-lang.org/download/ spark-1.4.0:必须是编译好的Spark,如果下载的是Source ...
- Windows下为64位的python3.4.3安装numpy
貌似现在没有python3.x的numpy 64位.exe安装包只有.whl的(也可能是我没找到)只能在终端下安装 1.到官网https://www.python.org/downloads/下载py ...
- URAL 1158 AC自动机上的简单DP+大数
题目大意 在一种语言中的字母表中有N(N<=50)个字母,每个单词都由M(M<=50)个字母构成,因此,一共可以形成N^M个单词.但是有P(P<=10)个串是被禁止的,也就是说,任何 ...
- poj 3264 Balanced Lineup (线段树)
Balanced Lineup Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 42489 Accepted: 20000 ...
- mybatis写mapper文件注意事项(转)
原文链接:http://wksandy.iteye.com/blog/1443133 xml中某些特殊符号作为内容信息时需要做转义,否则会对文件的合法性和使用造成影响 < < > & ...
- HTML--7JavaScript的DOM操作
1.DOM的基本概念 DOM是文档对象模型,这种模型为树模型:文档是指标签文档:对象是指文档中每个元素:模型是指抽象化的东西. 2.Window对象操作 一.属性和方法: 属性(值或者子对象): op ...
- C++11的new concepts (move semantic)
MoveConstructible 和MoveAssignable MoveConstructible Specifies that an instance of the type can be mo ...
- Java基础毕向东day02
1. 常量 null 等特殊 2.标识符 数字-字母-下划线,数字不能开头 3.二进制 1> 二进制计算方法. 2>常用二进制. 1 1 0 0 1 ...
- C杂记
printf主要是为了便于取地址,默认将char,short变成了int,float变成了double:但scanf的时候,因为数据宽度的问题,必须区分%f和%lf. memset 是字符串操作函数: ...