SGU 506 Subsequences Of Substrings

这个题 还是比较简单的；  把案例 想出来是怎么来的；差不多就出来了； 只要从前往后扫，遇到一个子序列就统计一次，更新起点；

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;

char str[112345],cha[112];
int main( )
{
    while( scanf("%s%s",str+1,cha+1) != EOF )
    {
        int len1 = strlen(str+1); int len2 = strlen(cha+1);
        int pre = 0; int now = 0; int k = 1; long long res = 0; int i = 1;
        while( i <= len1 )
        {
            if( str[i] == cha[k] )
            {
                if( k == 1 )now = i;
                if( k == len2 )
                {
                    res += (long long)(now-pre)*(long long)(len1-i+1);
                    k = 1; pre = now; i = now+1; now = 0;
                }else { i++; k++; }
            }else i++;
        }
        printf("%lld
",res);
    }
    return 0;
}