poj-1936-All in All
来源:互联网 发布:常用端口号列表 编辑:程序博客网 时间:2024/06/03 19:48
Description
You have devised a new encryption technique which encodes a message by inserting between its characters randomly generated strings in a clever way. Because of pending patent issues we will not discuss in detail how the strings are generated and inserted into the original message. To validate your method, however, it is necessary to write a program that checks if the message is really encoded in the final string.
Given two strings s and t, you have to decide whether s is a subsequence of t, i.e. if you can remove characters from t such that the concatenation of the remaining characters is s.
Given two strings s and t, you have to decide whether s is a subsequence of t, i.e. if you can remove characters from t such that the concatenation of the remaining characters is s.
Input
The input contains several testcases. Each is specified by two strings s, t of alphanumeric ASCII characters separated by whitespace.The length of s and t will no more than 100000.
Output
For each test case output "Yes", if s is a subsequence of t,otherwise output "No".
Sample Input
sequence subsequenceperson compressionVERDI vivaVittorioEmanueleReDiItaliacaseDoesMatter CaseDoesMatter
Sample Output
YesNoYesNo
字符串匹配 水题
#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>using namespace std; int main() { char str1[100005],str2[100005]; while(~scanf("%s%s",str1,str2)) { if(strstr(str2,str1)) { printf("Yes\n"); continue; } int len1,len2,i,j,cnt = 0,flag; len1 = strlen(str1); len2 = strlen(str2); for(i = 0; i<len1; i++) { flag = 1; for(j = 0; j<len2; j++) { if(str1[i] == str2[j]) { cnt++; flag = 0; strcpy(str2,str2+j+1); // puts(str2); len2 = strlen(str2); break; } } if(flag) { break; } } if(flag) { printf("No\n"); continue; } if(cnt == len1) printf("Yes\n"); else printf("No\n"); } return 0; }
0 0
- poj 1936 "all in all"
- poj 1936 All in All
- POJ 1936 All in All
- poj 1936 All in All
- POJ 1936 all in all
- POJ 1936 All in All
- POJ 1936 All in All
- POJ 1936 All in All
- POJ-1936-All in All
- poj 1936 all in all
- poj 1936 All in All
- POJ 1936 All in ALL
- poj 1936 All in All
- poj 1936 All in All
- POJ 1936 All in All
- All in All(poj 1936)
- poj 1936 All in All
- poj 1936 All in All
- 用C++编写小学生随机十道练习题的步骤以及源代码
- UVa 11997 K Smallest Sums (优先队列)
- FFmpeg深入分析之零-基础
- Javascript诞生记-C和Self语言的产物
- 1112. Stucked Keyboard (20) hash
- poj-1936-All in All
- 手机使用sideload升级失败 解决办法
- 彻底明白Android中AIDL及其使用
- CoreText NSTextView和Attribued String
- CAKeyframeAnimation实现抖动效果
- 关于Maven启动项目时总是下载POM的问题
- HttpClient4.5教程-第四章-HTTP身份认证
- 算法_动态规划_最少硬币问题
- [IO]——节点流