Give me the answer

Give me the answer

时间： 1ms        内存：32M

描述：

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)

输入：

The first line of the input contains the number of test cases in the file. And t he first line of each case

contains one integer numbers n

输出：

For each test case, output a line with the ans % 1000000000.

示例输入：

1
7

示例输出：

6

提示：

参考答案（内存最优[4908]）：

#include<stdio.h>
int main()
{
    int a[1000000];
    int n;
    scanf("%d",&n);
    while(n--)
   {
   int s;
   scanf("%d",&s);
    a[1]=1;
    int i;
    a[2]=2;
    a[3]=2;
    for(i=4;i<=s;i++)
    a[i]=(a[i-2]+a[i/2])%1000000000;
    printf("%d\n",a[s]%1000000000);
    }
    return 0;
    }

参考答案（时间最优[8]）：

#include <iostream>
using namespace std;
long long a[1000001];
int main()
{
    long long N,T,i;
    a[1]=1;
    a[2]=2;
    for(i=3; i<1000001; i++)
    {
        if(a[i]%2==1)
            a[i]=a[i-1];
        else
            a[i]=a[i-2]+a[i/2];
        a[i]%=1000000000;
    }
    cin>>N;
    while(N--)
    {
        cin>>T;
        cout<<a[T]<<endl;
    }
    return 0;
}

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