CF 163 A Substring and Subsequence（dp）

A. Substring and Subsequence

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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 ass = 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 sequencesp 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 test(s)

input

aa
aa

output

5

input

codeforces
forceofcode

output

60

Note

Let's write down all pairs "x y"
that form the answer in the first sample: "s[1...1] t[1]",
"s[2...2] t[1]",
"s[1...1] t[2]","s[2...2] t[2]",
"s[1...2] t[1 2]".

/*
思路:枚举 a[i]==b[j]的地方统计

*/

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define bug printf("hihi\n")
#define mod 1000000007
#define eps 1e-8
typedef __int64 ll;

using namespace std;
#define N 5005#define mod 1000000007

char a
,b
;
int lena,lenb;
ll dp

;

int main()
{
    int i,j;
    while(~scanf("%s%s",b+1,a+1))
    {
        lena=strlen(a+1);
        lenb=strlen(b+1);
        ll ans=0;
        for(i=1;i<=lena;i++)   //a 是不连续的
        {
            for(j=1;j<=lenb;j++)  //b是连续的
            {
               dp[i][j]=dp[i-1][j];   //dp[i][j]记录和前i-1和 j匹配的情况有多少种
               if(a[i]==b[j])  //当a[i]==b[j] 那么就是满足新的条件的
                  {
                      dp[i][j]=(dp[i][j]+dp[i-1][j-1]+1)%mod;
                      ans=(ans+dp[i-1][j-1]+1)%mod;
                  }
            }
        }
        printf("%I64d\n",ans);
    }
    return 0;
}