Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 863 Accepted Submission(s): 339
Problem 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 USACO 2005 January Silver
Recommend teddy 分析（转）：
设a[n]为和为 n 的种类数；
根据题目可知，加数为2的N次方，即 n 为奇数时等于它前一个数 n-1 的种类数 a[n-1] ，若 n 为偶数时分加数中有无 1 讨论，即关键是对 n 为偶数时进行讨论：
1.n为奇数，a[n]=a[n-1]
2.n为偶数：
（1）如果加数里含1，则一定至少有两个1，即对n-2的每一个加数式后面 +1+1，总类数为a[n-2]；
（2）如果加数里没有1，即对n/2的每一个加数式乘以2，总类数为a[n-2]；
所以总的种类数为：a[n]=a[n-2]+a[n/2];
#include<iostream> using namespace std; int a[1000000]; int main() { int i,n; a[1]=1; a[2]=2; for(i=3;i<=1000000;i++) { if(i%2==0) { a[i]=a[i-2]+a[i/2]; if(a[i]>999999999) a[i]%=1000000000; } else a[i]=a[i-1]; } while(~scanf("%d",&n)) { printf("%d\n",a[n]); } return 0; }
转载于:https://www.cnblogs.com/Su-Blog/archive/2012/09/11/2680751.html
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