题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5791
参考博客:https://blog.csdn.net/wuxuanyi27/article/details/52116674
Two
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2815 Accepted Submission(s): 1206
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
题目大意:给你两个集合,长度分别为n和m,需要你求出他们相同的子序列个数。
解题思路:看起来有点像最长公共子序列,不过有点不一样。我们可以很容易确定状态,用dp[i][j]表示第一个序列的前i个元素和第二个序列的前j个元素相同子序列的个数。关键是推导出状态方程,如果第一序列第i个元素和第二个序列的第j个元素不相同的话,我们需要考虑两种情况,如果第一个序列没有第i个元素,他们相同的子序列个数加上如果第二个序列没有第j个元素,他们相同子序列的个数,同时再减去dp[i-1][j-1],,因为在之前将第一个序列前i-1和第二个序列前j-1计算了两边,就可以的得到:dp[i][j]=dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1]。当第一个序列的第i个元素等于第二个序列的第j个元素的话,需要加上i与j相同的一个之外,还需要加上dp[i-1][j-1],因为dp[i-1][j-1]可以与i配对相同,也可以与j配对相同,于是就需要重复计算一次。
dp[i][j]=dp[i-1][j]+dp[i][j-1]+1。
dp[i][j]=dp[i-1][j]+dp[i][j-1]+1。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 using namespace std; 5 typedef long long ll; 6 const int MOD=1e9+7; 7 const int maxn=1005; 8 ll dp[maxn][maxn]; 9 int n,m; 10 int a[maxn],b[maxn]; 11 12 int main() 13 { 14 while(cin>>n>>m) 15 { 16 memset(dp,0,sizeof(dp)); 17 for (int i=1;i<=n;i++) 18 cin>>a[i]; 19 for (int i=1;i<=m;i++) 20 cin>>b[i]; 21 for(int i=1;i<=n;i++) 22 { 23 for(int j=1;j<=m;j++) 24 { 25 if(a[i]==b[j]) dp[i][j]=(dp[i-1][j]+dp[i][j-1]+1)%MOD; 26 else dp[i][j]=(dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1])%MOD; 27 } 28 } 29 cout<<(dp[n][m]+MOD)%MOD<<endl; 30 } 31 }