B - Sumsets 找规律 / 01背包

B - Sumsets


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


思路:前面几个数据比较小,可以写出来寻找规律。

T=1时,种类有1种(1);T=2时,种类有2种(1+1;2);当T=3时,种类有2种(1+1+1;1+2);当T=4时,种类有4种(1+1+1+1;1+1+2;2+2;4);当T=5时,种类有4种(1+1+1+1+1;1+1+1+2;1+2+2;1+4);

写到这里就不难发现,举个例子,当T=3的时候,3列出的数是在T=2种类数的基础上都加了1,得到的就是T=3时的种类数,

当T=4的时候,(2+2;4)是T=2时(1+1;2)的2倍,(1+1+1+1;1+1+2)是T=3时(1+1+1;1+2)每一个都多加1,


综上所述:

当为奇数的时候,种类数就和前面的偶数相同; a[i]=a[i-1]

当为偶数时,种类数就是后一个的种类数+二分之一的种类数; a[i]=a[i-1]+a[i/2]

#include
#include
#include
#include
using namespace std;
int g=1000000000;
int a[1000005];
int main(){
	int T,i,j;
	a[1]=1;
	a[2]=2;
	scanf("%d",&T);
	for(i=3;i<=T;i++){
		if(i%2!=0)
		a[i]=a[i-1];
		else
		a[i]=a[i-1]+a[i/2];
		a[i]=a[i]%g;
	}
	printf("%d\n",a[T]);
	return 0;
}


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