nyoj 308 Substring

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Substring

时间限制:1000 ms  |  内存限制:65535 KB
难度:1
描述

You are given a string input. You are to find the longest substring of input such that the reversal of the substring is also a substring of input. In case of a tie, return the string that occurs earliest in input. 

Note well: The substring and its reversal may overlap partially or completely. The entire original string is itself a valid substring . The best we can do is find a one character substring, so we implement the tie-breaker rule of taking the earliest one first.

输入
The first line of input gives a single integer, 1 ≤ N ≤ 10, the number of test cases. Then follow, for each test case, a line containing between 1 and 50 characters, inclusive. Each character of input will be an uppercase letter ('A'-'Z').
输出
Output for each test case the longest substring of input such that the reversal of the substring is also a substring of input
样例输入
3                   ABCABAXYZXCVCX
样例输出
ABAX

XCVCX

2015,4,17
最长公共子序列

#include<stdio.h>#include<string.h>char s[55],c[55];int a[55][55];int main(){int t,i,j,k,len,m;scanf("%d",&t);while(t--){m=0;memset(a,0,sizeof(a));scanf("%s",s);len=strlen(s);for(i=0,j=len-1;i<len;i++,j--)//翻转字符串 c[i]=s[j];for(i=1;i<=len;i++)for(j=1;j<=len;j++)if(s[i-1]==c[j-1]){a[i][j]=a[i-1][j-1]+1;//a[i][j]代表如果s[i-1]=c[j-1]的话,那么a[i][j]就等于a[i-1][j-1]+1; if(m<a[i][j]){//因为如果前两个不匹配,那么a[i-1][j-1]就是0,那么a[i][j]就是1;如果前边两个匹配 m=a[i][j];//那么a[i][j]就等于a[i-1][j-1]再加 1,就是2,a[i][j]就是指到ij的时候有几个字符匹配 k=i;//然后记录位置,输出就可以了,,这个过程如果还不懂就模拟一下,我也是看大神代码模拟一遍才理解 }}for(i=k-m;i<k;i++)printf("%c",s[i]);printf("\n");}return 0;}


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