Aaronson

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 231    Accepted Submission(s): 149

 

Problem Description

Recently, Peter saw the equation x0+2x1+4x2+...+2mxm=n. He wants to find a solution (x0,x1,x2,...,xm) in such a manner that ∑i=0mxi is minimum and every xi (0≤i≤m) is non-negative.

 

 

Input

There are multiple test cases. The first line of input contains an integer T (1≤T≤105), indicating the number of test cases. For each test case:

The first contains two integers n and m (0≤n,m≤109).

 

 

Output

For each test case, output the minimum value of ∑i=0mxi.

 

 

Sample Input

10 1 2 3 2 5 2 10 2 10 3 10 4 13 5 20 4 11 11 12 3

 

 

Sample Output

1 2 2 3 2 2 3 2 3 2

 

题目出处：中文翻译

 

从2的最大次幂开始除以保证最终得数最小。

 

附AC代码：

#include<iostream>
#include<cmath>
using namespace std;

/*
int Pow(int a,int b){
    int temp=1;
    for(int i=0;i<b;i++){
        temp*=a;
    }
    return temp;
}
*/

int main(){
    int t,m,n,k,sum;
    cin>>t;
    int s[31];
    s[0]=1;
    for(int i=1;i<31;i++){//打表 
        s[i]=s[i-1]+s[i-1];
    }
    while(t--){
        cin>>n>>m;
        sum=0;
        for(int j=min(m,30);j>=0;j--)  //当n可被除时，从最大2的最大次幂开始除以保证结果最小 
            if(n>=s[j])  
            {  
                k=n/s[j];  
                sum+=k;  
                n-=k*s[j];  
            }  
            cout<<sum<<endl;
    }
    return 0;
}