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
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2015,4,17最长公共子序列
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#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; }
时间: 2024-10-10 19:11:53