POJ 2229 递推

Posted 2021-01-25 kongbb

Farmer John commanded his cows to search for different sets of numbers that sum to a given number. The cows use only numbers that are an integer power of 2. Here are the possible sets of numbers that sum to 7: 

1) 1+1+1+1+1+1+1 
2) 1+1+1+1+1+2 
3) 1+1+1+2+2 
4) 1+1+1+4 
5) 1+2+2+2 
6) 1+2+4 

Help FJ count all possible representations for a given integer N (1 <= N <= 1,000,000). 

Input

A single line with a single integer, N.

Output

The number of ways to represent N as the indicated sum. Due to the potential huge size of this number, print only last 9 digits (in base 10 representation).

Sample Input

7

Sample Output

6

当 i 为奇数时可以将 i-1 的展开项加 1 得到 i 的展开项
当 i 为偶数是单纯的将 i-1 的展开项加 1 无法得到所有的 i 的展开项，因为 i 是偶数，所以 i 的展开项中有全为偶数的情况
将全为偶数的展开像除以 2 得到的是 i/2 的展开项（可以倒着想，将 i/2 的展开项乘 2 得到 i 的全偶数展开项） 

转移式为 dp[i] = (i&1)?(dp[i-1]):(dp[i-1]+dp[i/2])

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 using namespace std;
 5 
 6 const int maxn = 1e6+5;
 7 int dp[maxn];
 8 
 9 int main(){
10     int n;
11     scanf("%d",&n);
12     memset(dp,0,sizeof(dp));
13     dp[0] = 1;
14 
15     for(int i=1;i<=n;++i){
16         if(i&1)
17             dp[i] = dp[i-1];
18         else{
19             dp[i] = dp[i-1] + dp[i>>1];
20             if(dp[i]>1000000000)
21                 dp[i] -= 1000000000;
22         }
23     }
24 
25     printf("%d
",dp[n]);
26     return 0;
27 }

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