CodeForces 163A Substring and Subsequence dp A. Substring and Subsequence

题目连接：

http://codeforces.com/contest/163/problem/A

Description

One day Polycarpus got hold of two non-empty strings s and t, consisting of lowercase Latin letters. Polycarpus is quite good with strings, so he immediately wondered, how many different pairs of "x y" are there, such that x is a substring of string s, y is a subsequence of string t, and the content of x and y is the same. Two pairs are considered different, if they contain different substrings of string s or different subsequences of string t. Read the whole statement to understand the definition of different substrings and subsequences.

The length of string s is the number of characters in it. If we denote the length of the string s as |s|, we can write the string as s = s1s2... s|s|.

A substring of s is a non-empty string x = s[a... b] = sasa + 1... sb (1 ≤ a ≤ b ≤ |s|). For example, "code" and "force" are substrings or "codeforces", while "coders" is not. Two substrings s[a... b] and s[c... d] are considered to be different if a ≠ c or b ≠ d. For example, if s="codeforces", s[2...2] and s[6...6] are different, though their content is the same.

A subsequence of s is a non-empty string y = s[p1p2... p|y|] = sp1sp2... sp|y| (1 ≤ p1 < p2 < ... < p|y| ≤ |s|). For example, "coders" is a subsequence of "codeforces". Two subsequences u = s[p1p2... p|u|] and v = s[q1q2... q|v|] are considered different if the sequences p and q are different.

Input

The input consists of two lines. The first of them contains s (1 ≤ |s| ≤ 5000), and the second one contains t (1 ≤ |t| ≤ 5000). Both strings consist of lowercase Latin letters.

Output

Print a single number — the number of different pairs "x y" such that x is a substring of string s, y is a subsequence of string t, and the content of x and y is the same. As the answer can be rather large, print it modulo 1000000007 (109 + 7).

Sample Input

codeforces

forceofcode

Sample Output

60

Hint

题意

给你两个字符串，然后问你第一个字符串的子串和第二个子序列有多少对相同的串

题解：

dp[i][j]表示第一个串以i结尾，第二个串以j结尾的方案数

最简单的dp就是dp[i][j]=1+dp[i-1][j-1]+dp[i-1][j-2]+....+dp[i-1][1]

然后后面这个累加可以直接维护成前缀和就好了，这样复杂度就降下来了。

代码

#include<bits/stdc++.h>
using namespace std;

const int maxn = 5005;
const int mod = 1e9+7;
int dp[maxn][maxn];
char a[maxn],b[maxn];
int updata(int x,int y)
{
    return (x+y)%mod;
}
int main()
{
    scanf("%s%s",a+1,b+1);
    int len1 = strlen(a+1),len2 = strlen(b+1);
    for(int i=1;i<=len1;i++)
    {
        for(int j=1;j<=len2;j++)
        {
            if(a[i]==b[j])
                dp[i][j]=updata(dp[i][j],dp[i-1][j-1]+1);
            dp[i][j]=updata(dp[i][j],dp[i][j-1]);
        }
    }
    long long ans = 0;
    for(int i=1;i<=len1;i++)
        ans=updata(ans,dp[i][len2]);
    printf("%d
",ans);
}