zoukankan      html  css  js  c++  java
  • POJ 1936.All in All

    All in All
    Time Limit:1000MS     Memory Limit:30000KB     64bit IO Format:%I64d & %I64u

    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

    应该算是水题吧~

    用两个变量记录指向s、t字符串的位置。

    如果相等,则把指向s的+1

    看循环完s是否在最后即可

    AC代码:GitHub

     1 /*
     2 By:OhYee
     3 Github:OhYee
     4 HomePage:http://www.oyohyee.com
     5 Email:oyohyee@oyohyee.com
     6 Blog:http://www.cnblogs.com/ohyee/
     7 
     8 かしこいかわいい?
     9 エリーチカ!
    10 要写出来Хорошо的代码哦~
    11 */
    12 
    13 #include <cstdio>
    14 #include <algorithm>
    15 #include <cstring>
    16 #include <cmath>
    17 #include <string>
    18 #include <iostream>
    19 #include <vector>
    20 #include <list>
    21 #include <queue>
    22 #include <stack>
    23 #include <map>
    24 using namespace std;
    25 
    26 //DEBUG MODE
    27 #define debug 0
    28 
    29 //循环
    30 #define REP(n) for(int o=0;o<n;o++)
    31 
    32 const int maxn = 100005;
    33 
    34 bool Do() {
    35     char s[maxn],t[maxn];
    36     if(scanf("%s%s",s,t) == EOF)
    37         return false;
    38     
    39     int t_len = strlen(t);
    40     int s_len = strlen(s);
    41 
    42     int it = 0;
    43     for(int i = 0;i < t_len;i++) {
    44         if(s[it] == t[i])
    45             it++;
    46         if(it == s_len)
    47             break;
    48     }
    49 
    50     printf("%s
    ",(it == s_len) ? "Yes" : "No");
    51 
    52     return true;
    53 }
    54 
    55 int main() {
    56     while(Do());
    57     return 0;
    58 }
  • 相关阅读:
    微软Silverlight 2.0 最新版本GDR发布
    POJ 2635, The Embarrassed Cryptographer
    POJ 3122, Pie
    POJ 1942, Paths on a Grid
    POJ 1019, Number Sequence
    POJ 3258, River Hopscotch
    POJ 3292, Semiprime Hnumbers
    POJ 2115, C Looooops
    POJ 1905, Expanding Rods
    POJ 3273, Monthly Expense
  • 原文地址:https://www.cnblogs.com/ohyee/p/5472303.html
Copyright © 2011-2022 走看看