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![](http://acm.hdu.edu.cn/js/MathJax/fonts/HTML-CSS/TeX/png/Math/Italic/400/004E.png?rev=2.4-beta-2)
![](http://acm.hdu.edu.cn/js/MathJax/fonts/HTML-CSS/TeX/png/Main/Regular/400/002C.png?rev=2.4-beta-2)
![](http://acm.hdu.edu.cn/js/MathJax/fonts/HTML-CSS/TeX/png/Math/Italic/400/004D.png?rev=2.4-beta-2)
![](http://acm.hdu.edu.cn/js/MathJax/fonts/HTML-CSS/TeX/png/Main/Regular/400/0028.png?rev=2.4-beta-2)
![](http://acm.hdu.edu.cn/js/MathJax/fonts/HTML-CSS/TeX/png/Main/Regular/400/0031.png?rev=2.4-beta-2)
![](http://acm.hdu.edu.cn/js/MathJax/fonts/HTML-CSS/TeX/png/Main/Regular/400/2264.png?rev=2.4-beta-2)
![](http://acm.hdu.edu.cn/js/MathJax/fonts/HTML-CSS/TeX/png/Math/Italic/400/004E.png?rev=2.4-beta-2)
![](http://acm.hdu.edu.cn/js/MathJax/fonts/HTML-CSS/TeX/png/Main/Regular/400/002C.png?rev=2.4-beta-2)
![](http://acm.hdu.edu.cn/js/MathJax/fonts/HTML-CSS/TeX/png/Math/Italic/400/004D.png?rev=2.4-beta-2)
![](http://acm.hdu.edu.cn/js/MathJax/fonts/HTML-CSS/TeX/png/Main/Regular/400/2264.png?rev=2.4-beta-2)
![](http://acm.hdu.edu.cn/js/MathJax/fonts/HTML-CSS/TeX/png/Main/Regular/400/0031.png?rev=2.4-beta-2)
![](http://acm.hdu.edu.cn/js/MathJax/fonts/HTML-CSS/TeX/png/Main/Regular/400/0030.png?rev=2.4-beta-2)
![](http://acm.hdu.edu.cn/js/MathJax/fonts/HTML-CSS/TeX/png/Main/Regular/400/0030.png?rev=2.4-beta-2)
![](http://acm.hdu.edu.cn/js/MathJax/fonts/HTML-CSS/TeX/png/Main/Regular/400/0030.png?rev=2.4-beta-2)
. 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
![](http://acm.hdu.edu.cn/js/MathJax/fonts/HTML-CSS/TeX/png/Math/Italic/400/004E.png?rev=2.4-beta-2)
![](http://acm.hdu.edu.cn/js/MathJax/fonts/HTML-CSS/TeX/png/Main/Regular/400/002C.png?rev=2.4-beta-2)
![](http://acm.hdu.edu.cn/js/MathJax/fonts/HTML-CSS/TeX/png/Math/Italic/400/004D.png?rev=2.4-beta-2)
![](http://acm.hdu.edu.cn/js/MathJax/fonts/HTML-CSS/TeX/png/Main/Regular/400/0028.png?rev=2.4-beta-2)
![](http://acm.hdu.edu.cn/js/MathJax/fonts/HTML-CSS/TeX/png/Main/Regular/400/0031.png?rev=2.4-beta-2)
![](http://acm.hdu.edu.cn/js/MathJax/fonts/HTML-CSS/TeX/png/Main/Regular/400/2264.png?rev=2.4-beta-2)
![](http://acm.hdu.edu.cn/js/MathJax/fonts/HTML-CSS/TeX/png/Math/Italic/400/004E.png?rev=2.4-beta-2)
![](http://acm.hdu.edu.cn/js/MathJax/fonts/HTML-CSS/TeX/png/Main/Regular/400/002C.png?rev=2.4-beta-2)
![](http://acm.hdu.edu.cn/js/MathJax/fonts/HTML-CSS/TeX/png/Math/Italic/400/004D.png?rev=2.4-beta-2)
![](http://acm.hdu.edu.cn/js/MathJax/fonts/HTML-CSS/TeX/png/Main/Regular/400/2264.png?rev=2.4-beta-2)
![](http://acm.hdu.edu.cn/js/MathJax/fonts/HTML-CSS/TeX/png/Main/Regular/400/0031.png?rev=2.4-beta-2)
![](http://acm.hdu.edu.cn/js/MathJax/fonts/HTML-CSS/TeX/png/Main/Regular/400/0030.png?rev=2.4-beta-2)
![](http://acm.hdu.edu.cn/js/MathJax/fonts/HTML-CSS/TeX/png/Main/Regular/400/0030.png?rev=2.4-beta-2)
![](http://acm.hdu.edu.cn/js/MathJax/fonts/HTML-CSS/TeX/png/Main/Regular/400/0030.png?rev=2.4-beta-2)
![](http://acm.hdu.edu.cn/js/MathJax/fonts/HTML-CSS/TeX/png/Main/Regular/400/0029.png?rev=2.4-beta-2)
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
/* 给你两个序列,让你求有多少个公共子序列。 dp[i][j]表示第一串到i位置,第二串到j位置,最多有多少相同子序列; 那么状态转移方程就变成了dp[i][j]=dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1]; 这里列一个矩阵更好帮助理解 */ # include <iostream> # include <cstring> # include <stdio.h> using namespace std; const int MAX = 1005; const long long int mod = 1000000007; int a[MAX], b[MAX]; long long int dp[MAX][MAX]; int main() { int n, m; while(scanf("%d%d",&n,&m)) { 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)); for(int i = 1; i <= n; i++) for(int j = 1; j <= m; j++) { if(a[i] == b[j]) dp[i][j] = (dp[i - 1][j] + dp[i][j - 1] + 1) % mod; else dp[i][j] = (dp[i - 1][j] + dp[i][j - 1] - dp[i - 1][j - 1] + mod)%mod;//注意这里很可能出现负数 } printf("%d ",dp[n][m]); } return 0; }