欢迎您访问程序员文章站本站旨在为大家提供分享程序员计算机编程知识!
您现在的位置是: 首页  >  IT编程

DP_Sumsets

程序员文章站 2022-03-21 21:36:20
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 ......

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

 

题意:整数n用2^n之和的形式表示的方案数

思路:当n>1时,若n为奇数,则每个分解方案中至少含有一个1项。此时若每种分解方案中去掉一个1项,方案数不发生改变。

   即     n分解的方案数=(n-1)分解的方案数

   若该n为偶数,则分为两类情况:

   1、含有1项,同上,每个方案中去掉1项,方案数不变。

   2、不含有1项,此时每个方案中最小项应为2,若将每一项除2,方案数不变。

   即     n分解的方案数=(n-1)分解的方案数+(n/2)分解的方案数。

边界条件:当n=1时只有一种分解方案。

注意:可能溢出,需要取模

 

 

 1 #include<cstdio>
 2 int s[1000005];
 3 int main()
 4 {
 5     int n;
 6 
 7     s[1]=1;
 8     for( int i=2; i<=1000000; i++){
 9         if( i%2==0 )
10             s[i]=(s[i-1]+s[i/2])%1000000000;
11         else
12             s[i]=s[i-1];
13     }
14     while(~scanf("%d",&n)){
15         printf("%d\n",s[n]);
16     }
17 
18     return 0;
19 }