poj2229Sumsets

http://poj.org/problem?id=2229

挺好的一公式。。

 1 #include <iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<stdlib.h>
 6 using namespace std;
 7 #define  N 1000010
 8 #define mod 1000000000
 9 long long dp[N];
10 int a[N];
11 int main()
12 {
13     int i,j,k,n;
14     while(cin>>n)
15     {
16         memset(dp,0,sizeof(dp));
17         memset(a,0,sizeof(a));
18         dp[1] = 1;
19         dp[2] = 2;
20         dp[3] = 2;
21         dp[4] = 4;
22         for(i = 5; i <= n ; i++)
23         {
24             if(i%2!=0)
25             dp[i] = dp[i-1];
26             else
27             {
28                 dp[i] = (dp[i-2]+dp[i/2])%mod;
29             }
30         }
31         cout<<dp[n]%mod<<endl;
32     }
33     return 0;
34 }
View Code
原文地址:https://www.cnblogs.com/shangyu/p/3251696.html