cf.VK CUP 2015.C.Name Quest(贪心）

Name Quest

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

A Martian boy is named s — he has got this name quite recently from his parents for his coming of age birthday. Now he enjoys looking for his name everywhere. If he sees that he can obtain his name from some string by removing zero or more letters (at that, the remaining letters remain in the same order), he gets happy. For example, if s=«aba», then strings «baobab», «aabbaa», «helloabahello» make him very happy and strings «aab», «baaa» and «helloabhello» do not.

However rather than being happy once, he loves twice as much being happy twice! So, when he got string t as a present, he wanted to cut it in two parts (the left part and the right part) so that each part made him happy.

Help s determine the number of distinct ways to cut the given string t into two parts in the required manner.

Input

The first line contains string s, consisting of lowercase English letters. The length of string s is from 1 to 1000 letters.

The second line contains string t, that also consists of lowercase English letters. The length of string t is from 1 to 10⁶ letters.

Output

Print the sought number of ways to cut string t in two so that each part made s happy.

Sample test(s)

input

aba baobababbah

output

2

input

mars sunvenusearthmarsjupitersaturnuranusneptune

output

0
先从左到右，找到子串中的第一个母串，让l = sub[index];（对应母串的最后一个字母）
再从右往左找第一个符合的母串，让r = sub[index] ;(对应母串的第一个字母）
最后printf ("%d
" , max (0 , r - l))
然后我把代码写搓了 ， orz

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
string p , s ;

int main ()
{
    //freopen ("a.txt" , "r" , stdin ) ;
    cin >> p ;
    cin >> s ;
    int l = 0 , r = s.size () - 1;
    for (int i = 0 ; i < p.size() ; i++) {
        while (l < s.size () && p[i] != s[l]) {
            l ++ ;
        }
        if (l < s.size ()) {
            l ++ ;
        }
        else {
            puts ("0") ;
            return 0 ;
        }
    }
    for (int i = p.size () - 1 ; i >= 0 ; i--) {
        while (r >= 0 && p[i] != s[r]) {
            r-- ;
        }
        if (r >= 0) {
            r-- ;
        }
        else {
            puts ("0") ;
            return 0 ;
        }
    }
    l -- , r++ ;
    printf ("%d
" , max (0 , r - l) ) ;
}