Two
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1533 Accepted Submission(s): 676Problem 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 2Sample 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:
#include<bits/stdc++.h>
using namespace std; const int pr=; int dp[][];
int a[],b[]; int main(){
int n,m;
while(cin>>n>>m){
for(int i=;i<=n;i++){
for(int j=;j<=m;j++){
dp[i][j]=;
}
}
for(int i=;i<=n;i++){
cin>>a[i];
}
for(int i=;i<=m;i++){
cin>>b[i];
}
for(int i=;i<=n;i++){
for(int j=;j<=m;j++){
dp[i][j]=dp[i][j-]+dp[i-][j]-dp[i-][j-];
if(a[i]==b[j])
dp[i][j]+=dp[i-][j-]+;
if(dp[i][j]<)
dp[i][j]+=pr;
if(dp[i][j]>=pr)
dp[i][j]%=pr;
}
}
cout<<dp[n][m]<<endl;
}
return ;
}
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