atcoder #027 B. Garbage Collector Author: xiaotianyao

版权声明：转载请注明出处 https://blog.csdn.net/qq_41593522/article/details/84196549

题意

把n个垃圾捡回0，捡起一个花费x，回一次0花费x，行走1格花费(身上垃圾数 + 1)^2

问最小花费

题解

反正要捡n个，花费n * x，先不管这个

设分成k次捡，则每组中最远的到最近的对答案的贡献分别 * 5, 5, 7, …

枚举k，计算答案，求出最优解

调试记录

sum若已经 >= ans，要及时退出，防止溢出

#include <cstdio>
#include <algorithm>
#define maxn 200005
#define int long long
#define INF 0x3f3f3f3f3f3f3f3f

using namespace std;

int n, a[maxn], x;

signed main(){
	scanf("%lld%lld", &n, &x);
	a[0] = 0;
	for (int i = 1; i <= n; i++) scanf("%lld", a + i), a[i] += a[i - 1];
	
	int ans = INF, sum;
	for (int k = 1; k <= n; k++){
		sum = x * n + x * k; int t = 3;
		for (int i = n; i >= 1; i -= k){
			sum += (a[i] - a[max(0ll, i - k)]) * max(5ll, t), t += 2;
			if (sum >= ans) break;
		}
		ans = min(ans, sum);
	} printf("%lld\n", ans);
	
	return 0;
}