Two
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1533 Accepted Submission(s): 676
Problem DescriptionAlice 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.
InputThe 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.
OutputFor each test case, output the answer mod 1000000007.
Sample Input3 2 1 2 3 2 1 3 2 1 2 3 1 2
Sample Output2 3
题意:
求公共子序列数量。
dp[i][j]表示第一个串考虑到i位,第二个串考虑到j位的答案是多少。
那么dp[i][j] = dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1] ,需要特别判断a[i]=b[j]时,dp[i][j]+=dp[i-1][j-1]+1。
附AC代码:
1 #include<bits/stdc++.h> 2 using namespace std; 3 4 const int pr=1000000007; 5 6 int dp[2010][2010]; 7 int a[2010],b[2010]; 8 9 int main(){ 10 int n,m; 11 while(cin>>n>>m){ 12 for(int i=1;i<=n;i++){ 13 for(int j=1;j<=m;j++){ 14 dp[i][j]=0; 15 } 16 } 17 for(int i=1;i<=n;i++){ 18 cin>>a[i]; 19 } 20 for(int i=1;i<=m;i++){ 21 cin>>b[i]; 22 } 23 for(int i=1;i<=n;i++){ 24 for(int j=1;j<=m;j++){ 25 dp[i][j]=dp[i][j-1]+dp[i-1][j]-dp[i-1][j-1]; 26 if(a[i]==b[j]) 27 dp[i][j]+=dp[i-1][j-1]+1; 28 if(dp[i][j]<0) 29 dp[i][j]+=pr; 30 if(dp[i][j]>=pr) 31 dp[i][j]%=pr; 32 } 33 } 34 cout<<dp[n][m]<<endl; 35 } 36 return 0; 37 }