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

1 2

3 2

5 2

10 2

10 3

10 4

13 5

20 4

11 11

12 3

Sample Output

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;

}

巴特西

hdu-5747 Aaronson(水题)

Aaronson

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