hdu 5747 Aaronson【贪心】思维的体操

Recently, Peter saw the equation x0+2x1+4x2+...+2mxm=nx0+2x1+4x2+...+2mxm=n . He wants to find a solution (x0,x1,x2,...,xm)(x0,x1,x2,...,xm) in such a manner that ∑i=0mxi∑i=0mxi is minimum and every xixi (0≤i≤m0≤i≤m ) is non-negative.

Input

There are multiple test cases. The first line of input contains an integer TT (1≤T≤105)(1≤T≤105) , indicating the number of test cases. For each test case:

The first contains two integers nn and mm (0≤n,m≤109)(0≤n,m≤109) .

Output

For each test case, output the minimum value of ∑i=0mxi∑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

思路：关注方程左边x前的系数是不断上升的，有贪心可出答案（一开始没思路，就是没有观察到方程的性质特点）

#include<cstdio>
#include<math.h>
#include<string.h>
#include<vector>
#include<algorithm>
#include<queue>
#include<bitset>
#include<iostream>

#define pi pair<int,int> 
#define CLR(a,b) memset(a,b,sizeof(a))
#define ll long long

using namespace std;
const int maxn=1e4+10;
ll a[32];

void init(void){
	a[0]=1;
	for(int i=1;i<=31;i++){
		a[i]=a[i-1]*2;
	}
}

int main(){
	int t;
	init();
	while(~scanf("%d",&t)){
		int n,m;
		while(t--){
			ll ans=0;
			scanf("%d%d",&n,&m);
			for(int i=31;i>=0;i--){
				if(n/a[i]>0){
					ans+=n/a[i];
					n%=a[i];
				}
			}
			printf("%lld\n",ans);
		}
	}
}