题目链接：

Aaronson

Time Limit: 4000/2000 MS (Java/Others)    

Memory Limit: 131072/131072 K (Java/Others)

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

题意：

把n拆成这样,系数和最小,贪心;

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
//#include <bits/stdc++.h>
#include <stack>

using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));

typedef  long long LL;

template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<‘0‘||CH>‘9‘;F= CH==‘-‘,CH=getchar());
    for(num=0;CH>=‘0‘&&CH<=‘9‘;num=num*10+CH-‘0‘,CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + ‘0‘);
    putchar(‘\n‘);
}

const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=1e5+10;
const int maxn=500+10;
const double eps=1e-8;

LL f[60];
void Init()
{
    f[0]=1;
    For(i,1,50)
    {
        f[i]=f[i-1]*2;
    }
}

int main()
{       

        int t;
        read(t);
        Init();
        while(t--)
        {
         int n,m;
         read(n);read(m);
         int ans=0;
         for(int i=min(40,m);i>=0;i--)
         {
            if(n>=f[i])
            {
                ans+=n/f[i];
                n=n%f[i];
            }
         }
         printf("%d\n",ans);
        }
        return 0;
}