http://acm.hdu.edu.cn/showproblem.php?pid=5791

Two

Problem Description

Alice gets two sequences A and B. A easy problem comes. How many pair of sequence A‘ and sequence B‘ are same. For example, {1,2} and {1,2} are same. {1,2,4} and {1,4,2} are not same. A‘ is a subsequence of A. B‘ is a subsequence of B. The subsequnce can be not continuous. For example, {1,1,2} has 7 subsequences {1},{1},{2},{1,1},{1,2},{1,2},{1,1,2}. The answer can be very large. Output the answer mod 1000000007.

Input

The input contains multiple test cases.

For each test case, the first line cantains two integers N,M(1≤N,M≤1000). The next line contains N integers. The next line followed M integers. All integers are between 1 and 1000.

Output

For each test case, output the answer mod 1000000007.

Sample Input

3 2

1 2 3

2 1

3 2

1 2 3

1 2

Sample Output

2

3

题意：有两个串，求两两子串相同的个数有多少（可以不连续）。

思路：有点类似于LCS的DP。

 1 #include <cstdio>
 2 #include <algorithm>
 3 #include <cstring>
 4 using namespace std;
 5 #define N 1005
 6 #define MOD 1000000007
 7 typedef long long LL;
 8
 9 LL dp[N][N];
10 int a[N], b[N];
11 /*
12 1 2 3
13 2 1
14 */
15 int main()
16 {
17     int n, m;
18     while(~scanf("%d%d", &n, &m)) {
19         dp[0][0] = 0;
20         for(int i = 1; i <= n; i++) {
21             scanf("%d", a+i);
22             dp[i][0] = 0;
23         }
24         for(int i = 1; i <= m; i++) {
25             scanf("%d", b+i);
26             dp[0][i] = 0;
27         }
28 /*
29 有这三部分
30 dp[i-1][j-1]
31 dp[i-1][j] - dp[i-1][j-1]
32 dp[i][j-1] - dp[i-1][j-1]
33 如果不匹配的话 dp[i][j] = dp[i-1][j] - dp[i-1][j-1] + dp[i][j-1] - dp[i-1][j-1] + dp[i-1][j-1]
34 匹配的话 dp[i][j] = 不匹配的状态 + dp[i-1][j-1] + 1
35 dp[i-1][j] + dp[i][j-1] - dp[i-1][j-1] 表示当前不匹配的状态有多少种，因为dp[i-1][j]和dp[i][j]中有dp[i-1][j-1]重复，所以要减去一个
36 如果当前匹配的话，就不用减去，因为要留一个来和当前的a[i]和b[j]匹配。
37 */
38         for(int i = 1; i <= n; i++) {
39             for(int j = 1; j <= m; j++) {
40                 if(a[i] == b[j]) {
41                     dp[i][j] = (dp[i-1][j] + dp[i][j-1] + 1 + MOD) % MOD;
42                 } else {
43                     dp[i][j] = (dp[i-1][j] + dp[i][j-1] - dp[i-1][j-1] + MOD) % MOD;
44                 }
45             }
46         }
47
48         printf("%I64d\n", dp[n][m] % MOD);
49     }
50     return 0;
51 }