Description
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
Source
如果i为奇数，肯定有一个1，把f[i-1]的每一种情况加一个1就得到fi,所以f[i]=f[i-1]
如果i为偶数，如果有1，至少有两个，则f[i-2]的每一种情况加两个1，就得到i，如果没有1，则把分解式中的每一项除2，则得到f[i/2]
所以f[i]=f[i-2]+f[i/2]
由于只要输出最后9个数位，别忘记模1000000000
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #include<cmath> 7 using namespace std; 8 #define N 1010006 9 #define MOD 1000000000 10 int n; 11 int dp[N]; 12 void init(){ 13 14 dp[1]=1; 15 dp[2]=2; 16 for(int i=3;i<N;i++){ 17 if(i&1){ 18 dp[i]=dp[i-1]; 19 } 20 else{ 21 dp[i]=dp[i-1]+dp[i/2]; 22 dp[i]%=MOD; 23 } 24 } 25 } 26 int main() 27 { 28 init(); 29 while(scanf("%d",&n)==1){ 30 31 printf("%d ",dp[n]); 32 33 } 34 return 0; 35 }