子序列（All in all), UVa 10340

#include <iostream>
#include <algorithm>
#include <string>
#include <cstring>
using namespace std;
int main()
{
	string s, t;
	while (cin >> s && cin >> t)
	{
		int lent = s.length();
		int is_sub = 1;
		long long currentpos = 0;
		for (int i = 0; i < lent; ++i)
		{
			if ((currentpos = t.find(s[i], currentpos)) == string::npos)
			{
				is_sub = 0;
				break;
			}
			++currentpos;//t数组中索引后移一位
		}
		if (is_sub == 0)
			cout << "No" << endl;
		else
			cout << "Yes" << endl;
	}
}