糖果传递 BZOJ1045

糖果传递 BZOJ1045

思路

题干在这:BZOJ1045
就是前缀和+排序+中位数,没什么好说的qwq

ac代码

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
long long a[1000010] = { 0 };
long long s[1000010] = { 0 };
int main() {
	int n;
	long long sum = 0;
	cin >> n;
	for (int i = 1; i <= n; i++) {
		cin >> a[i];
		sum += a[i];
	}
	long long average = sum / n;
	for (int i = 1; i <= n; i++) {
		a[i]-=average;
		s[i] = s[i - 1] + a[i];
	}
	sort(s + 1, s + n + 1);
	long long sum1 = 0, sum2 = 0;
	for (int i = 1; i <= n; i++) {
		sum1 += abs(s[i] - s[n / 2 + 1]);
		sum2 += abs(s[i] - s[n / 2]);
	}
	long long resort = min(sum1, sum2);
	cout << resort;
}
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转载自blog.csdn.net/qq_45616764/article/details/104203009