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  • Palindrome subsequence

                    Palindrome subsequence

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65535 K (Java/Others)
    Total Submission(s): 2232    Accepted Submission(s): 889


    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 <A, B, D> is a subsequence of <A, B, C, D, E, F>.
    (http://en.wikipedia.org/wiki/Subsequence)

    Given a string S, your task is to find out how many different subsequence of S is palindrome. Note that for any two subsequence X = <Sx1, Sx2, ..., Sxk> and Y = <Sy1, Sy2, ..., Syk> , if there exist an integer i (1<=i<=k) such that xi != yi, the subsequence X and Y should be consider different even if Sxi = Syi. Also two subsequences with different length should be considered different.
     
    Input
    The first line contains only one integer T (T<=50), which is the number of test cases. Each test case contains a string S, the length of S is not greater than 1000 and only contains lowercase letters.
     
    Output
    For each test case, output the case number first, then output the number of different subsequence of the given string, the answer should be module 10007.
     
    Sample Input
    4
    a
    aaaaa
    goodafternooneveryone
    welcometoooxxourproblems
     
    Sample Output
    Case 1: 1
    Case 2: 31
    Case 3: 421
    Case 4: 960
     
    状态方程dp[i][j] = dp[i+1][j]+dp[i][j-1] - dp[i+1][j-1]; 如果s[i] ==s[j] , dp[i][j]还要加上dp[i+1][j-1]+1; 
    这道题WA了很惨,自己做题太少,对于有 -号再求余的一定要考虑是否有可能得出负数,加上mod之后可以保证是正数 dp[i][j] = (dp[i+1][j]+dp[i][j-1] - dp[i+1][j-1] +mod)%mod;
     
     1 #include <iostream>
     2 #include <algorithm>
     3 #include <cstring>
     4 #include <cstdio>
     5 using namespace std;
     6 const int N = 1005;
     7 const int mod = 10007;
     8 int _, n, dp[N][N], cas=1;
     9 char s[N];
    10 
    11 void solve()
    12 {
    13     scanf("%s", s+1);
    14     n = strlen(s+1);
    15     memset(dp, 0, sizeof(dp));
    16     for(int k=0; k<n; k++)
    17     {
    18         for(int i=1; i+k<=n; i++)
    19         {
    20             int t = i+k;
    21             dp[i][t] = (dp[i+1][t] + dp[i][t-1] - dp[i+1][t-1] + mod) % mod;/////////注意
    22             if(s[i]==s[t]) dp[i][t] += dp[i+1][t-1] + 1;
    23             dp[i][t] %= mod;
    24         }
    25     }
    26     printf("Case %d: %d
    ", cas++, dp[1][n]);
    27 }
    28 
    29 int main()
    30 {
    31     scanf("%d", &_);
    32     while(_--) solve();
    33     return 0;
    34 }
    View Code
     
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  • 原文地址:https://www.cnblogs.com/zyx1314/p/3843323.html
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