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;
}