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  • nyoj 308 Substring

    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                   

    ABCABA

    XYZ

    XCVCX

    样例输出

    ABA

    X

    XCVCX

    #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("
    ");
    	}
    	return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/zhangliu/p/7057727.html
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