Bestcoder-Aaronson

Aaronson

 

 Accepts: 632

 

 Submissions: 1442

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

 

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

Problem Description

Recently, Peter saw the equation x_{0}+2x_{1}+4x_{2}+...+2^{m}x_{m}=nx​0​​+2x​1​​+4x​2​​+...+2​m​​x​m​​=n. He wants to find a solution (x_0,x_1,x_2,...,x_m)(x​0​​,x​1​​,x​2​​,...,x​m​​) in such a manner that \displaystyle\sum_{i=0}^{m} x_i​i=0​∑​m​​x​i​​ is minimum and every x_ix​i​​ ($0\le i\le m$) is non-negative.

Input

There are multiple test cases. The first line of input contains an integer $T$ (1 \le T \le 10^5)(1≤T≤10​5​​), indicating the number of test cases. For each test case:

The first contains two integers $n$ and $m$ (0 \le n,m \le 10^9)(0≤n,m≤10​9​​).

Output

For each test case, output the minimum value of \displaystyle\sum_{i=0}^{m} x_i​i=0​∑​m​​x​i​​.

Sample Input

101 23 25 210 210 310 413 520 411 1112 3

Sample Output

1223223232

简单的二进制题

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<limits.h>#include<math.h>#include<algorithm>using namespace std;int  main(){int t,n,m;int i,j;scanf("%d",&t);while (t--){scanf("%d%d",&n,&m);j=n;int sum=0,x;for(i=1;i<=m;i++){sum+=n%2;n=n/2;}sum+=n;printf("%d\n",sum);}}