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CodeForces 558C Amr and Chemistry (位运算,数论,规律,枚举)

Codeforces 558C

题意:给n个数字,对每一个数字能够进行两种操作:num*2与num/2(向下取整),求:让n个数相等最少须要操作多少次。

分析:

计算每一个数的二进制公共前缀.

枚举法亦可。

/*
*Author : Flint_x 
*Created Time : 2015-07-22 12:33:11 
*File name : whust2_L.cpp 
*/

#include<iostream>
#include<sstream>
#include<fstream>
#include<vector>
#include<list>
#include<deque>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cctype>
#include<cmath>
#include<ctime>
#include<iomanip>
#define inf 2139062143
using namespace std;
const double eps(1e-8);

typedef long long lint;

#define cls(a) memset(a,0,sizeof(a))
#define rise(i,a,b) for(int i = a ; i <= b ; i++)
#define fall(i,a,b) for(int i = a ; i >= b ; i--)

const int maxn = 100000 + 5;
int num[maxn];
int temp[maxn];
int odd[maxn],cnt[maxn];
int n;

int main(){
   // freopen("input.txt","r",stdin);
//  freopen("output.txt","w",stdout);
	
	while(cin >> n){
		for(int i = 1  ; i <= n ; i++){
			scanf("%d",&num[i]);
		}
		cls(cnt);cls(odd);
		sort(num+1,num+n+1);
		for(int i = 1 ; i <= n ; i++) temp[i] = num[i];
		int t = num[1];
		for(int i = 1 ; i <= n ; i++){
			while(t ^ num[i]){
				if (t < num[i]) num[i] >>= 1;
				else t >>= 1;
			}
		}
		for(int i = 1 ; i <= n ; i++) num[i] = temp[i];
		for(int i = 1 ; i <= n ; i++){
			while (num[i] ^ t){
				cnt[i]--;
				if(num[i] % 2) odd[i] = cnt[i];
				num[i] >>= 1;
			}
		}
		lint ans = inf;
		for(int i = 0 ; i < 20 ; i++){
			lint x = 0;
			for(int j = 1 ; j <= n ; j++){
				if (odd[j] == 0 || cnt[j] + i <= odd[j]) x += abs(cnt[j] + i);
				else x += abs(odd[j]) + abs(odd[j] - (cnt[j] + i));
				
			}
//			cout << x << endl;
			ans = min(ans,x);
		}
		cout << ans << endl;
	}
    return 0;
}
        


CodeForces 558C Amr and Chemistry (位运算,数论,规律,枚举)