Two
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 2692 Accepted Submission(s): 1156
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 21 2 32 13 21 2 31 2
Sample Output
23
题意:给定两个数列。求出它们公共子序列的数量并对1e9+7取模
一开始没什么思路,后面看了大佬们的代码发现可以用dp做,emmm
切入正题,我们可以用dp[i][j]表示A序列到i位,B序列到j位相同子序列的个数
那么a[i]==b[j]的话
dp[i][j]=(dp[i-1][j]+d[i][j-1]+1)%mod;
不满足的话(这里要考虑防止相减得出负数的情况,所以加上mod)
dp[i][j]=(dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1]+mod)%mod;
附上全部代码:
#include <iostream> #include <cstdio> #define mod 1000000007 using namespace std; int a[1005],b[1005]; long long dp[1005][1005]; int main(){ int n,m; // freopen("in.in","r",stdin); while(cin>>n>>m){ dp[0][0]=dp[0][1]=dp[1][0]=0; for(int i=1;i<=n;i++) cin>>a[i]; for(int i=1;i<=m;i++) cin>>b[i]; 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; } } } cout<<dp[n][m]<<endl; } }