Two
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1421 Accepted Submission(s): 630
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
1 2 3
2 1
3 2
1 2 3
1 2
Sample Output
2
3
3
求两个序列的相同的公共子序列的个数，取模。
DP[i][j]表示到a数组的第 i 个和b数组的第 j 个之间的个数，可以归纳出，当a[i]==b[j]时，dp[i][j]=dp[i-1][j]+dp[i][j-1]+1;
当不等的时候，dp[i][j]=dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1].(仔细琢磨琢磨，会发现很有道理)因为有减法的运算所以最后答案可能出现
负数，注意处理一下。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std; typedef long long ll;
const int mod = ;
ll dp[][];
int a[],b[]; int main()
{
int n,m;
while (~scanf("%d%d",&n,&m)){
for (int i= ; i<=n ; i++) scanf("%d",&a[i]);
for (int i= ; i<=m ; i++) scanf("%d",&b[i]);
memset(dp,,sizeof(dp));
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-]+;
else dp[i][j]=dp[i-][j]+dp[i][j-]-dp[i-][j-];
dp[i][j]%=mod;
}
}
printf("%I64d\n",(dp[n][m]+mod)%mod);
}
return ;
}