Codeforce 163 A. Substring and Subsequence DP

[b]A. Substring and Subsequence[/b]

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.

[b]Input[/b]
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.

[b]Output[/b]
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).

[b]Sample test(s)[/b]

input

aa
aa

output

5

input

codeforces
forceofcode

output

60

[b]Note[/b]
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]".

[b]题意： [/b]

　　给出两个串，问a的子串和b的子序列（可以不连续）相同的个数。

[b]题解：[/b]

　　dp[i][j]以a[i]结尾的以b[j]结尾相同个数。

　　那么 if(a[i] == a[j]) dp[i][j] = dp[i-1][j-1] + dp[i-1][j-2] + ............+dp[i-1][1] + 1

此时当做前缀和处理就可以了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;
typedef long long ll;
const int  N = 5000 + 10;
const int mod = 1e9 + 7;
ll dp

;
int main() {
char a
,b
;
scanf("%s%s",a+1,b+1);
int len = strlen(a+1);
int lenb = strlen(b+1);
for(int i = 1; i <= len; i++) {
for(int j = 1; j <= lenb; j++) {
dp[i][j] = dp[i][j-1] % mod;
if(a[i] == b[j]) dp[i][j] = (dp[i][j] + dp[i-1][j-1] + 1) % mod;
}
}
ll ans = 0;
for(int i = 1; i <= len; i++) ans = (ans + dp[i][lenb]) %mod;
printf("%I64d\n",ans);
return 0;
}