BestCoder Round #84 1001 hdu 5747 模拟

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链接：戳这里

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
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：

直接将n去填(1<<i)就可以了，还有就是m不能超过32

代码：

#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>#define mst(ss,b) memset((ss),(b),sizeof(ss))#define maxn 0x3f3f3f3f#define MAX 1000100///#pragma comment(linker, "/STACK:102400000,102400000")typedef long long ll;typedef unsigned long long ull;#define INF (1ll<<60)-1using namespace std;int a[33];int n,m;int main(){    for(int i=0;i<=30;i++) a[i]=(1<<i);    int T;    scanf("%d",&T);    while(T--){        scanf("%d%d",&n,&m);        int x=lower_bound(a,a+30,n)-a;        int ans=0;        for(int i=min(m,x);i>=0;i--){            ans=ans+n/a[i];            n%=a[i];            if(n==0) break;        }        printf("%d\n",ans);    }    return 0;}

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