HDU 5791 Two （DP）

HDU 5791 Two （DP）

Two

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 540    Accepted Submission(s): 242

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

Author

ZSTU

Source

​​2016 Multi-University Training Contest 5​​

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题意：给你a,b两个数组，问你有多少个公共子序列。

题解：DP。

设dp[ i ][ j ]表示a中前i个和b中前j个并且a[i]==b[j]匹配的子序列个数。

那么dp[ i ][ j ] = dp[ i-1 ][ j ]+dp[ i ][ j-1 ]-dp[ i-1 ][ j-1 ]。

当a[ i ]=b[ j ]时，dp[ i ][ j ]+=dp[ i-1 ][ j-1 ]+1

注意dp转移时要特殊处理一下。这题就可以了。

AC代码：

#includeusing namespace std;const int N=1010;typedef long long LL;const long long mod=1000000007;LL dp[N][N],sum1[N][N],sum2[N][N];int a[N],b[N];int main(){ int n,m; while(~scanf("%d%d",&n,&m)) { for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) dp[i][j]=0; for(int i=1;i<=n;i++) scanf("%d",&a[i]); for(int i=1;i<=m;i++) scanf("%d",&b[i]); for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) { dp[i][j]=dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1]; if(a[i]==b[j]) dp[i][j]+=dp[i-1][j-1]+1; if(dp[i][j]<0) dp[i][j]+=mod; //注意要加 mod，因为上面的dp有可能会减成负数 if(dp[i][j]>=mod) dp[i][j]%=mod; } } printf("%I64d\n",dp[n][m]); } return 0;}

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