POJ 2229 Sumsets(规律)
Posted shuaihui
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这是一道意想不到的规律题。。。。。。。。。。。。或许是我比较菜,找不到把。
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).
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
规律是:1 2 2 4 4 6 6 10 10 14 14 20 20 26 26 36,,,由这里不难可以看出,如果是奇数位,则于他前一位相当。。。重点是偶数位,d[i]=d[i-1]+d[i/2],这规律真的是。。。。
AC代码:
#define MOD 10000000000 #define MAX 20000 #include<stdio.h> int d[MAX]; int main() { int n; scanf("%d",&n); d[1]=1; d[2]=2; for(int i=3 ;i<=n ;i++) if(i & 1) d[i]=d[i-1]%MOD; else d[i]=(d[i-1]+d[i/2])%MOD; printf("%d\n",d[n]); return 0; }
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