HDU 4632 Palindrome subsequence(区间DP)
2016-03-04 08:14
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Palindrome subsequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65535 K (Java/Others)
Total Submission(s): 2830 Accepted Submission(s): 1156
Problem Description
In mathematics, a subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements. For example, the sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65535 K (Java/Others)
Total Submission(s): 2830 Accepted Submission(s): 1156
Problem Description
In mathematics, a subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements. For example, the sequence
// // main.cpp // 区间dp 1001 // // Created by 陈永康 on 16/2/28. // Copyright © 2016年 陈永康. All rights reserved. // #include <iostream> #include <string.h> #include <stdlib.h> #include <math.h> #include <algorithm> using namespace std; char a[1005]; int dp[1005][1005]; int main() { int t; scanf("%d",&t); int cas=0; while(t--) { scanf("%s",a+1); int len=strlen(a+1); memset(dp,0,sizeof(dp)); for(int i=1;i<=len;i++) dp[i][i]=1; for(int l=1;l<len;l++) { for(int i=1;i+l<=len;i++) { int j=i+l; // if(a[i]==a[j]) // { // if(i==j-1) // dp[i][j]+=3; // else // dp[i][j]=dp[i+1][j-1]*2+3; // // } // else // { // if(i==j-1) // dp[i][j]+=2; // else // dp[i][j]=dp[i+1][j-1]+2; // } // for(int k=i+1;k<j;k++) // { // if(a[i]==a[k]) // { // if(i==k-1) // dp[i][j]++; // else // dp[i][j]+=(dp[i+1][k-1])+1; // } // if(a[k]==a[j]) // { // if(k==j-1) // dp[i][j]++; // else // dp[i][j]+=(dp[k+1][j-1])+1; // } // } dp[i][j]=(dp[i+1][j]+dp[i][j-1]-dp[i+1][j-1]+10007)%10007; if(a[i]==a[j]) dp[i][j]=(dp[i][j]+dp[i+1][j-1]+1+10007)%10007; //dp[i][j]%=10007; } } printf("Case %d: %d\n",++cas,dp[1][len]); } return 0; }
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