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  • HDU 5642 King's Order(数位dp)

    King's Order




    Problem Description
    After the king's speech , everyone is encouraged. But the war is not over. The king needs to give orders from time to time. But sometimes he can not speak things well. So in his order there are some ones like this: "Let the group-p-p three come to me". As you can see letter 'p' repeats for 3 times. Poor king!

    Now , it is war time , because of the spies from enemies , sometimes it is pretty hard for the general to tell which orders come from the king. But fortunately the general know how the king speaks: the king never repeats a letter for more than 3 times continually .And only this kind of order is legal. For example , the order: "Let the group-p-p-p three come to me" can never come from the king. While the order:" Let the group-p three come to me" is a legal statement.

    The general wants to know how many legal orders that has the length of n

    To make it simple , only lower case English Letters can appear in king's order , and please output the answer modulo 1000000007

    We regard two strings are the same if and only if each charactor is the same place of these two strings are the same.
     
    Input
    The first line contains a number T(T10)——The number of the testcases.

    For each testcase, the first line and the only line contains a positive number n(n2000).
     
    Output
    For each testcase, print a single number as the answer.
     
    Sample Input
    2
    2
    4
     
    Sample Output
    676
    456950
     
     
    hint: All the order that has length 2 are legal. So the answer is 26*26. For the order that has length 4. The illegal order are : "aaaa" , "bbbb"…….."zzzz" 26 orders in total. So the answer for n == 4 is 26^4-26 = 456950
     
     1 #include<iostream>
     2 #include<cstdio>
     3 using namespace std;
     4 const int MOD=1e9+7;
     5 typedef long long ll;
     6 ll dp[2005][4];
     7 int main()
     8 {
     9     dp[1][1]=26;
    10     for(int i=2;i<=2000;i++)
    11     {
    12         dp[i][1]=(dp[i-1][1]+dp[i-1][2]+dp[i-1][3])*25%MOD;
    13         dp[i][2]=dp[i-1][1]%MOD;
    14         dp[i][3]=dp[i-1][2]%MOD;
    15     }
    16     int T;
    17     scanf("%d",&T);
    18     while(T--)
    19     {
    20         int n;
    21         scanf("%d",&n);
    22         printf("%d
    ",(dp[n][1]+dp[n][2]+dp[n][3])%MOD);
    23     }
    24     return 0;
    25 }
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  • 原文地址:https://www.cnblogs.com/homura/p/5390583.html
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