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.
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：
给你两个数组
问你有多少对公共子序列
题解：
设定dp[i][j] 表示以i结尾 j结尾的子序列的 答案数
n^2的转移
假设当前为 a[i] == b[j] , 那么它可以继承的 就是 所有 的 i,j组合 +1
a[i] != b[j] 则 当前dp[i][j] = 0 咯
第一中继承 只需要利用前缀 优化就行
最后统计答案的话 就是sum[n][m]咯
#include<iostream> #include<cstring> #include<cstdio> #include<cmath> #include<algorithm> using namespace std; const int N = 1e3+10, M = 2e2+11, inf = 2e9, mod = 1e9+7; typedef long long ll; ll dp[N][N],sum[N][N],ans = 0; int a[N],b[N],n,m; int main() { while (scanf("%d%d", &n, &m)!=EOF) { for (int i=1;i<=n;i++) scanf("%d", &a[i]); for (int i=1;i<=m;i++) scanf("%d", &b[i]); memset(dp,0,sizeof(dp)); memset(sum,0,sizeof(sum)); for (int i=1;i<=n;i++) for (int j=1;j<=m;j++) if (a[i]==b[j]) { dp[i][j]=(sum[i-1][j-1]+1)%mod; sum[i][j]=(-sum[i-1][j-1]+dp[i][j]%mod+sum[i-1][j]%mod+sum[i][j-1])%mod; } else sum[i][j]=(-sum[i-1][j-1]%mod+sum[i-1][j]%mod+sum[i][j-1])%mod; ans=0; for (int i=1;i<=n;i++) for (int j=1;j<=m;j++) ans=(ans+dp[i][j])%mod; printf("%I64d ", (ans+mod)%mod); } return 0; }