nyoj 308 Substring
来源:互联网 发布:王震对新疆的功过知乎 编辑:程序博客网 时间:2024/03/29 22:47
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;}
0 0
- NYOJ 308 Substring
- NYOJ 308题 Substring
- nyoj-308-Substring
- nyoj 308 Substring
- nyoj 308 Substring
- nyoj 308 Substring
- NYOJ 308-Substring【模拟】
- NYOJ 308 Substring
- nyoj 308 Substring
- NYOJ Substring--308
- nyoj 308 Substring
- nyoj 308 Substring
- NYOJ 308 Substring
- NYOJ—308—Substring
- NYOJ - Substring
- NYOJ 308 Substring 字符串处理问题
- nyoj 308 substring (最长逆序字符串)
- nyoj 308-Substring(find 和 substr的用法)
- FutureTask
- Java多线程总结之线程安全队列Queue
- 第一个 OpenGL ES的小程序(过程写的非常详细)
- C++ STL基本容器的使用
- Android-------模拟用户登录界面(5)
- nyoj 308 Substring
- Linux 生成可以调试的动态库(单步调试)
- 使用Dijkstra迪杰斯特拉算法获得单源最短路径
- MYSQL [Warning] Aborted connection 42355 to db
- eclipse juno(4.2) 集成 maven 插件
- Scala 学习笔记(4)-集合类简单操作
- python函数转模块命名
- 将ecshop中的session机制重写,从DB移植到Memcache中去
- 为什么牛顿法下降的速度比梯度下降的快