HDU 5791：Two（DP）

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，（+MOD）%MOD是因为有可能减出负数，因为取MOD一开始可能很大，后面变得很小。

 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 using namespace std;
 #define N 1005
 #define MOD 1000000007
 typedef long long LL;

 LL dp[N][N];
 int a[N], b[N];
 /*
 1 2 3
 2 1
 */
 int main()
 {
     int n, m;
     while(~scanf("%d%d", &n, &m)) {
         dp[][] = ;
         for(int i = ; i <= n; i++) {
             scanf("%d", a+i);
             dp[i][] = ;
         }
         for(int i = ; i <= m; i++) {
             scanf("%d", b+i);
             dp[][i] = ;
         }
 /*
 有这三部分
 dp[i-1][j-1]
 dp[i-1][j] - dp[i-1][j-1]
 dp[i][j-1] - dp[i-1][j-1]
 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]
 匹配的话 dp[i][j] = 不匹配的状态 + dp[i-1][j-1] + 1
 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
 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。
 */
         for(int i = ; i <= n; i++) {
             for(int j = ; j <= m; j++) {
                 if(a[i] == b[j]) {
                     dp[i][j] = (dp[i-][j] + dp[i][j-] +  + MOD) % MOD;
                 } else {
                     dp[i][j] = (dp[i-][j] + dp[i][j-] - dp[i-][j-] + MOD) % MOD;
                 }
             }
         }

         printf("%I64d\n", dp[n][m] % MOD);
     }
     return ;
 }