Two

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 21    Accepted Submission(s): 10

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

 类似最长公共子序列的dp。

#include <bits/stdc++.h> using namespace std; #define ll long long const int maxn = 1e5 + 10; ll a[maxn], b[maxn]; ll sum[maxn]; const ll mod = 1000000007; ll dp[1010][1010]; int main() { int n, m; while(~scanf("%d %d", &n, &m)) { memset(dp, 0, sizeof(dp)); for(int i = 1; i <= n; i++) scanf("%I64d", &a[i]); for(int i = 1; i <= m; i++) scanf("%I64d", &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]) % mod - dp[i - 1][j - 1] + mod) + (a[i] == b[j] ? dp[i - 1][j - 1] + 1 : 0)) % mod; } } printf("%I64d\n", dp[n][m] % mod); } }

 

转载于:https://www.cnblogs.com/lonewanderer/p/5730065.html