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  • POJ1936-All in All

    题目链接:点击打开链接

    All in All

    Time Limit: 1000MS   Memory Limit: 30000K
    Total Submissions: 35348   Accepted: 14736

    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.

    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 subsequence
    person compression
    VERDI vivaVittorioEmanueleReDiItalia
    caseDoesMatter CaseDoesMatter
    

    Sample Output

    Yes
    No
    Yes
    No
    

    题目大意:和第一题一样,在长的里面找短的所有字符。

    思路:刚开始没有这样写,用的find,然后比较下标,但是由于find是第一个找到的下标,出现重复字母就错了,后来就用第一种了。

    AC代码:

    #include<iostream>
    
    #include<string>
    #include<cstdio>
    using namespace std;
    
    string s, t;
    int main() {
    	while(cin >> s) {
    		cin >> t;
    
    		int j = 0, i;
    		for(i = 0; i < s.size(); ) {//跑一遍,找到字符就往后,找不到就不动
    			if(j == t.size())
    				break;
    			if(t[j] == s[i]) {
    				j++;
    				i++;
    			} else {
    				j++;
    			}
    		}
    		if(i == s.size())//判断是否能跑完
    			printf("Yes
    ");
    		else
    			printf("No
    ");
    	}
    }
    
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  • 原文地址:https://www.cnblogs.com/ACMerszl/p/9572973.html
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