hdu5747 Aaronson 贪心

Aaronson

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 425    Accepted Submission(s): 255

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

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<vector>
#include <ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<iomanip>
#include<cmath>
using namespace std;
int main(){
    int t;
    int a[32];
    for(int i=0;i<31;i++){
        a[i]=(1<<i);
    }
    scanf("%d",&t);
    while(t--){
        int n,m;
        int ans=0;
        scanf("%d%d",&n,&m);
        int x =lower_bound(a,a+30,n)-a;
        for(int i=min(x,m);i>=0;i--){
       
            ans+=(n/a[i]);
            n%=a[i];
            if(n==0) break;
        }
        printf("%d
",ans);
    }
    return 0;
}