YTU 2896: J--Zipper

2896: J--Zipper

时间限制: 1 Sec 内存限制: 128 MB

提交: 29 解决: 15

题目描述

Given three strings, you are to determine whether the third string can be formed by combining the characters in the first two strings. The first two strings can be mixed arbitrarily, but each must stay in its original order.

For example, consider forming "tcraete" from "cat" and "tree":

String A: cat

String B: tree

String C: tcraete

As you can see, we can form the third string by alternating characters from the two strings. As a second example, consider forming "catrtee" from "cat" and "tree":

String A: cat

String B: tree

String C: catrtee

Finally, notice that it is impossible to form "cttaree" from "cat" and "tree".

输入

The first line of input contains a single positive integer from 1 through 1000. It represents the number of data sets to follow. The processing for each data set is identical. The data sets appear on the following lines, one data set per line.

For each data set, the line of input consists of three strings, separated by a single space. All strings are composed of upper and lower case letters only. The length of the third string is always the sum of the lengths of the first two strings. The first two
strings will have lengths between 1 and 200 characters, inclusive.

输出

For each data set, print:

Data set n: yes

if the third string can be formed from the first two, or

Data set n: no

if it cannot. Of course n should be replaced by the data set number. See the sample output below for an example.

样例输入

3

cat tree tcraete

cat tree catrtee

cat tree cttaree

样例输出

Data set 1: yes

Data set 2: yes

Data set 3: no

你离开了，我的世界里只剩下雨。。。

#include<stdio.h>

#include<string.h>

#define MAX 201

int opt[MAX][MAX];

int main()

{

    int n;

    int cnt=0;

    char str1[MAX],str2[MAX],str3[MAX*2];

    int len_str1,len_str2,len_str3;

    int i,j;

    scanf("%d",&n);

    while(n--)

    {

        cnt++;

        scanf("%s %s %s",str1,str2,str3);

        len_str1=strlen(str1);

        len_str2=strlen(str2);

        len_str3=strlen(str3);

        if(len_str1+len_str2!=len_str3)

        {

            printf("Data set %d: no\n",cnt);

            continue;

        }

        memset(opt,0,sizeof(opt));

        opt[0][0]=1;

        for(i=1; i<=len_str1; i++)

        {

            if(opt[i-1][0]==1&&str1[i-1]==str3[i-1]) opt[i][0]=1;

            else  break;

        }

        for(j=1; j<=len_str2; j++)

        {

            if(opt[0][j-1]==1&&str2[j-1]==str3[j-1]) opt[0][j]=1;

            else  break;

        }

        for(i=1; i<=len_str1; i++)

        {

            for(j=1; j<=len_str2; j++)

            {

                if((opt[i][j-1]==1&&str2[j-1]==str3[i+j-1])||(opt[i-1][j]==1&&str1[i-1]==str3[i+j-1]))

                    opt[i][j]=1;

            }

        }

        if(opt[len_str1][len_str2])  printf("Data set %d: yes\n",cnt);

        else printf("Data set %d: no\n",cnt);

    }

    return 0;

}

巴特西

YTU 2896: J--Zipper

2896: J--Zipper

题目描述

输入

输出

样例输入

样例输出

你离开了，我的世界里只剩下雨。。。

最新文章

热门文章

巴特西

YTU 2896: J--Zipper

2896: J--Zipper

题目描述

输入

输出

样例输入

样例输出

你 离 开 了 ， 我 的 世 界 里 只 剩 下 雨 。 。 。

最新文章

热门文章

你离开了，我的世界里只剩下雨。。。