hdu 5747 Aaronson（贪心）

Aaronson

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 173    Accepted Submission(s): 106

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

101 23 25 210 210 310 413 520 411 1112 3

 

Sample Output

1223223232

题意：

x0+2x1+4x2+...+2mxm=n  给m和m。求最小的x序列之和

思路：10的9次方最多是2的30次方，我们贪心的每次从最高位取到不能取，最后的和一定是最小的

因为是二进制所以一定可以拼出n

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;long long pow_mod(long long a,long long n){    long long ans=1;    while(n)    {        if(n&1) ans*=a;        a*=a;        n>>=1;    }    return ans;}int main(){    int T;    long long n,m;    scanf("%d",&T);    while(T--)    {        scanf("%lld %lld",&n,&m);        long long ans=0;        if(m>30) m=30;        while(n)        {            long long t=n/pow_mod(2,m);            ans+=t;            n-=t*pow_mod(2,m);            m--;        }        printf("%d\n",ans);    }    return 0;}