描述

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.

输入

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.

输出

For each test case output "Yes", if s is a subsequence of t,otherwise output "No".

样例输入

sequence subsequence
person compression
VERDI vivaVittorioEmanueleReDiItalia
caseDoesMatter CaseDoesMatter

样例输出

Yes
No
Yes
No

解法:

#include<stdio.h>
#include<string.h>
#define MAXN 100000
char first[MAXN+];
char second[MAXN+];
int main(void)
{
int first_len, second_len, i, j;
bool find;
while(scanf("%s%s", first, second) != EOF)
{
first_len = strlen(first);
second_len = strlen(second);
if(first_len > second_len)
printf("No\n");
else
{
j = ;
for(i = ; i < first_len; i++)
{
find = false; // 主要用来应付最后一个字符
for(; j < second_len; j++)
{
if(second[j] == first[i])
{
j++;
find = true;
break;
}
} if(j == second_len)
break;
}
if(find && i >= first_len-)
printf("Yes\n");
else
printf("No\n");
}
} return ;
}

开始的时候,Yes 和No 都写成了大写,一直WA,郁闷了个数小时。。。
更简洁的写法:

#include<stdio.h>
#include<string.h>
#define MAXN 100000
char first[MAXN+];
char second[MAXN+];
int main(void)
{
int i, j, first_len, second_len; while(scanf("%s%s", first, second) != EOF)
{
first_len = strlen(first);
second_len = strlen(second);
i = j = ;
while(i < first_len && j < second_len)
{
if(first[i] == second[j])
{
i++;
j++;
}
else
j++; } if(i == first_len)
printf("Yes\n");
else
printf("No\n");
} return ;
}

OpenJudge_1936:All in All的更多相关文章

随机推荐

  1. Javascript-简单的计时钟表

    <!DOCTYPE html> <html lang="en" xmlns="http://www.w3.org/1999/xhtml"> ...

  2. zeroclipboard复制插件兼容IE8

    <!DOCTYPE html> <html> <head lang="en"> <meta charset="UTF-8&quo ...

  3. 基于jquery实现图片上传本地预览功能

    一.原理 分为两步: 当上传图片的input被触发并选择本地图片之后获取要上传的图片这个对象的URL(对象URL),把对象URL赋值给事先写好的img标签的src属性即可把图片显示出来.在这里,我们需 ...

  4. shell总结:读取文件、参数、if、分割字符串、数组长度、空文件、变量赋值、多进程、按行切割文件、查看线程

    Reference: http://saiyaren.iteye.com/blog/1943207 1.     Shell  读取文件和写文件 for line in $(<top30000. ...

  5. 备忘 ubuntu ip 及 dns 的坑

    以前都用 ubuntu 16.04 现在用 18.04 遇到几个恶心的事,现在解决了,记录下来. 1. 设置 DNS  ,    DNS 设置老是不对,最后发现问题老版本 ubuntu 17.10以下 ...

  6. jnhs中国的省市县区邮编坐标mysql数据表

    https://blog.csdn.net/sln2432713617/article/details/79412896 -- 1.之前项目中需要全国的省市区数据,在网上找了很多,发现数据要么不全,要 ...

  7. paip.前端载入时间分析之道优化最佳实践

    paip.前端载入时间分析之道优化最佳实践 1.另存为 ,查看文件尺寸..和图片. 2.view the 另存为的htm静态的文件单个的载入,看时间...能够排除编程语言的问题and 数据库.. ## ...

  8. Leetcode40. Combination Sum组合总数2 II

    给定一个数组 candidates 和一个目标数 target ,找出 candidates 中所有可以使数字和为 target 的组合. candidates 中的每个数字在每个组合中只能使用一次. ...

  9. 精密MRAM芯片制造系统

    MRAM是一种非常复杂的薄膜多层堆叠,由10多种不同材料和超过30层以上的薄膜与堆叠组成,部分薄膜层的厚度仅达数埃,比人类的发丝还要薄500000倍,相近于一颗原子的大小,如何控制这些薄膜层的厚度.沉 ...

  10. Linux操作系统各版本ISO镜像下载(包括oracle linux\redhat\centos\u

    Linux操作系统各版本ISO镜像下载(包括oracle linux\redhat\centos\ubuntu\debian等) 1.Oracle Linux(下载地址) (1)OracleLinux ...