POJ 2681
#include<iostream>
#include<stdio.h>
#include<string>
#include<algorithm>
#define MAXN 50
using namespace std;
char ch1[MAXN];
char ch2[MAXN];
int dig1[MAXN];
int dig2[MAXN];
int main()
{
//freopen("acm.acm","r",stdin);
int num;
char c;
int i;
int j;
int ans;
int len[];
int time;
int place;
int begin;
cin>>num;
getchar();
for(time = ; time <= num; ++ time)
{
ans = ;
memset(dig1,-,sizeof(dig1));
memset(dig2,-,sizeof(dig2));
gets(ch1);
gets(ch2);
len[] = strlen(ch1);
len[] = strlen(ch2);
for(i = ; i < len[]; ++ i)
{
dig1[i] = ch1[i] - 'a';
}
for(i = ; i < len[]; ++ i)
{
dig2[i] = ch2[i] - 'a';
}
//sort(dig2,dig2+len[1]);
//sort(dig1,dig1+len[0]);
begin = ;
for(i = ;i < len[]; ++ i)
{
for(j = ; j < len[]; ++ j)
if(ch1[i] == ch2[j])
{
ch2[j] = '~';
++ ans;
break;
}
}
cout<<"Case #"<<time<<": "<<len[]+len[]-*ans<<endl;
ch1[] = '\0';
ch2[] = '\0';
}
}
POJ 2681的更多相关文章
- poj 2681 字符串
http://poj.org/problem?id=2681 给你任意长度的字符串,找出两串字符中不相同的字符个数(总数) #include<string> #include<cst ...
- POJ 3370. Halloween treats 抽屉原理 / 鸽巢原理
Halloween treats Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 7644 Accepted: 2798 ...
- POJ 2356. Find a multiple 抽屉原理 / 鸽巢原理
Find a multiple Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 7192 Accepted: 3138 ...
- POJ 2965. The Pilots Brothers' refrigerator 枚举or爆搜or分治
The Pilots Brothers' refrigerator Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 22286 ...
- POJ 1753. Flip Game 枚举or爆搜+位压缩,或者高斯消元法
Flip Game Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 37427 Accepted: 16288 Descr ...
- POJ 3254. Corn Fields 状态压缩DP (入门级)
Corn Fields Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 9806 Accepted: 5185 Descr ...
- POJ 2739. Sum of Consecutive Prime Numbers
Sum of Consecutive Prime Numbers Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 20050 ...
- POJ 2255. Tree Recovery
Tree Recovery Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 11939 Accepted: 7493 De ...
- POJ 2752 Seek the Name, Seek the Fame [kmp]
Seek the Name, Seek the Fame Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 17898 Ac ...
随机推荐
- [转]ajQuery的deferred对象详解
来自:http://www.ruanyifeng.com/blog/2011/08/a_detailed_explanation_of_jquery_deferred_object.html 作者: ...
- 关于上级机构的冲突性测试bug修复
描述: 1.上级机构可以为空. 2.机构添加时,选择了上级机构,在未提交前,另一用户将该机构删除,然后前一用户再提交表单,提示会保存成功,本操作应该保存失败. 思路:在上级机构不为空时,保存前进行查询 ...
- POSTMAN 数据关联
概述 在使用postman测试接口是,我们可能需要先获取一个token,然后再将这个token发送到第二个请求.这个需要做postman的关联,一次性完成这两个测试. 实现方法 1.编写两个控制器方法 ...
- Docker Compose demo 使用
1.docker compose 安装 curl -L "https://github.com/docker/compose/releases/download/1.22.0/docker- ...
- Amaze UI 云适配
Amaze UI 云适配 陈本峰 一中,中科大 香港科大
- How to transfer developer profile to one mac to another mac
Export developer profile from old mac. In the Xcode Organizer, select your team in the Teams section ...
- (数学)Knight's Trip -- hdu -- 3766
http://acm.hdu.edu.cn/showproblem.php?pid=3766 Knight's Trip Time Limit: 2000/1000 MS (Java/Others) ...
- php获取跳转后的真实链接
网站的跳转链接经常为本站链接加上一些参数来跳转,如何使用php获取跳转后的链接呢? php代码如下: <?php // echo get_redirect_url('http://www.osc ...
- shell工具-cut
cut cut的工作就是“剪”,具体说就是在文件中负责剪切数据用的.cut命令从文件的每一行剪切字节.字符.和字段并将这些字节.字符和字段输出 基本用法 cut [参数] filename # 说明: ...
- PCB中实现元器件旋转一个角度放置
我们常常放置器件都是横着或者竖着的...但是有时候需要器件能旋转一个角度放更方便的话,可以这样 设置器件的属性.....