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[];
     s[]=;
     for(int i=;i<;i++){//打表
         s[i]=s[i-]+s[i-];
     }
     while(t--){
         cin>>n>>m;
         sum=;
         for(int j=min(m,);j>=;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 ;
 }