Codeforces A. Substring and Subsequence

链接：

A. Substring and Subsequence

大意：

找一个字符串的substring子串与另一个字符串的subsequence子序列相等的个数

思路：DP

代码：

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 2e5 + 10, INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;


void solve()
{
	string a, b;cin >> a >> b;
	int n = a.size(), m = b.size();
	//dp[i][j]是以i结尾的substring和以j结尾的subsequence
	vector<vector<int>>dp(n + 1, vector<int>(m + 1)); 
	int ans = 0;
	for (int i = 1; i <= n; i++)
	{
		for (int j = 1; j <= m; j++)
		{
			dp[i][j] = dp[i][j - 1]; //默认是以i结尾的字串和前j-1个子序列的方案数
			if (a[i - 1] == b[j - 1])
				dp[i][j] = (dp[i][j] + dp[i - 1][j - 1] + 1) % mod; // 方案数为（i-1,j-1）加上一种相同的尾
		}
		ans = (ans + dp[i][m]) % mod;
	}
	cout << ans << endl;
}

signed main()
{
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	int t = 1;
	//cin >> t;
	while (t--) solve();

	return 0;
}