2021_CCPC_harbin_I. Power and Zero

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本文介绍了一道编程竞赛题，题目要求通过一系列特定操作将一个正整数数组变为全零数组，每次操作可以对任意元素进行2的幂次减1的操作。解题思路涉及二分查找和位运算，通过示例解析了如何找到最少的操作次数。

传送门:https://codeforces.com/gym/103447/problem/I
二分+二进制拆分
题目大意：给定一个数组，每次操作可以选择任意个元素（可重复选择），对这多个元素分别 -1，-2，-4，-8…。
求最少操作多少次使得数组变为全0数组。
I. Power and Zero
time limit per test1 second
memory limit per test512 megabytes
inputstandard input
outputstandard output
Given a sequence A1,A2,⋯,An whose elements are all positive integers. You should do some operations to make the sequence all zero. For each operation, you can appoint a sequence B1,B2,⋯,Bm(Bi∈{1,2,⋯,n}) of arbitrary length m and reduce ABi by 2i−1 respectively. Specially, one element in the given sequence can be reduced multiple times in one operation. Determine the minimum possible number of operations to make the given sequence all zero.

Input
The first line contains one integer T(1≤T≤1000), denoting the number of test cases.

For each test case:

The first line contains one integer n(1≤n≤1e5), denoting the length of given sequence.

The second line contains n integers A1,A2,⋯,An(1≤Ai≤1e9), denoting the given sequence.

It is guaranteed that ∑n≤1e5.

Output
For each test case:

Output one line containing one integer, denoting the answer.

Example
inputCopy
3
5
1 2 3 4 5
2
1 4
1
7
outputCopy
3
3
1
Note
For the first sample case, one possible scheme:

Appoint B={1,3,5}, then A will be {0,2,1,4,1}.
Appoint B={3,2,4}, then A will be {0,0,0,0,1}.
Appoint B={5}, then A will be {0,0,0,0,0}.
For the second sample case, one possible scheme:

Appoint B={1,2}, then A will be {0,2}.
Appoint B={2}, then A will be {0,1}.
Appoint B={2}, then A will be {0,0}.
For the third sample case, one possible scheme:

Appoint B={1,1,1}, then A will be {0}.

#include<bits/stdc++.h>
using namespace std;
using ll = long long;
ll ay[40];
int n;
bool check(ll x){
	if(ay[0]>x) return 0;
	ll a[40];
	for(int i=0;i<=31;i++){
		a[i]=ay[i];
	}
	a[32]=0;
	for(int i=0;i<=31;i++){
		if(i&&a[i]>a[i-1]||a[i]>x) return false;
		ll y = i?a[i-1]-a[i]:x-a[i];
		a[i+1]-=y/2;
		a[i]+=y/2*2;
	}
	return true;
}
ll find()
{
	ll l=0 , r=1e16;
    while (l < r)
    {
        ll mid = l + r >> 1;
        if (check(mid)) r = mid;
        else l = mid + 1;
    }
    return l;
}
int main()
{
	int _;scanf("%d",&_);
	while(_--){
		memset(ay,0,sizeof ay);
		scanf("%d",&n);
		for(int i=1;i<=n;i++){
			ll x;scanf("%lld",&x);
			for(int j=0;j<=31;j++){
				ay[j]+=( (x>>j)&1 );
			}
		}
		printf("%lld\n",find());
	}
	
	
	return 0;
}