<LightOJ 1338> Hidden Secret!

Hidden Secret!

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

Submit Status Practice LightOJ 1338

Description

In this problem you are given two names, you have to find whether one name is hidden into another. The restrictions are:

1.      You can change some uppercase letters to lower case and vice versa.

2.      You can add/remove spaces freely.

3.      You can permute the letters.

And if two names match exactly, then you can say that one name is hidden into another.

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case starts with two lines. Each line contains a name consists of upper/lower case English letters and spaces. You can assume that the length of any name is between 1 and 100 (inclusive).

Output

For each case, print the case number and "Yes" if one name is hidden into another. Otherwise print "No".

Sample Input

3

Tom Marvolo Riddle

I am Lord Voldemort

I am not Harry Potter

Hi Pretty Roar to man

Harry and Voldemort

Tom and Jerry and Harry

Sample Output

Case 1: Yes

Case 2: Yes

Case 3: No

思路很简单，用整型数组a[]记录第一个name中出现的字母和个数，再跟之后输入的第二个name比较，具体实现方法见程序，此题也可以用STL map。

//Memory: 904 KBTime: 0 MS//Language: CResult: Accepted#include <stdio.h>#include <memory.h>#include <string.h>int main(){    int T;    int a[26], cnt1, cnt2;//分别记录两个name的长度(除space)    char c;    int flag;    scanf("%d ", &T);//%d后留一个空格就能忽略输入T时输入的回车    for(int ca=1; T--; ca++)    {        flag=1;        cnt1 = cnt2 = 0;        memset(a, 0, sizeof(a));        while(scanf("%c", &c) && c != '\n')        {//putchar(c);            if(c == ' ') continue;            if(c <= 'Z') a[c - 'A']++;            else a[c - 'a']++;            cnt1++;        }        //for(int i=0; i<26; i++) printf("%d ", a[i]);        while(scanf("%c", &c) && c != '\n')        {            if(c == ' ' || !flag) continue;            if(c <= 'Z')            {                if(a[c - 'A'])                    a[c - 'A']--;                else                    flag=0;            }            else            {                if(a[c - 'a'])                    a[c - 'a']--;                else                {                    flag=0;                    //break;//不可break，因为输入没结束！血淋淋的WA。。。                }            }            cnt2++;        }        //注意此处判断，这是针对第二个name是第一个name的非连续子序列的情况        if(cnt1 != cnt2) flag = 0;        if(flag) printf("Case %d: Yes\n", ca);        else printf("Case %d: No\n", ca);    }    return 0;}