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;
}

题目和答案均来自于互联网,仅供参考,如有问题请联系管理员修改或删除。

点赞

发表评论

电子邮件地址不会被公开。必填项已用 * 标注