AGC027 B - Garbage Collector 枚举/贪心

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目录

• [TOC]

题目链接

AGC027 B - Garbage Collector

题解

对于一组选取组的最优方案为，走到一点，然后顺着路径往回取点

设选取点坐标升序为{a,b,c,d}

那么消耗为\(d+(d - c) + 4* (d - c) + 9 * (c - d) + 16 * (b - a) + a * 25\)

化简后为\(5d + 5c + 7b - 9a\),那个对于这组k的最优解显然是让最远的点系数最小

考虑把序列划分为n / k组,枚举这个k计算,那么复杂度是调和级数的

代码

#include<cstdio>
#include<cstring>
#include<algorithm>
#define gc getchar()
#define pc putchar
#define int long long
inline int read() {
	int x = 0,f = 1;
	char c = gc;
	while(c < '0' || c > '9') {if(c == '-')f = -1; char c = getchar();}
	while(c <= '9' && c >= '0') x = x * 10 + c - '0',c = getchar();
	return x * f;
}
void print(long long  x) {
	if(x < 0) {
		pc('-');
		x = -x;
	}
	if(x >= 10) print(x / 10);
	pc(x % 10 + '0');
}
const int maxn = 500007;
#define LL long long
LL ans;
LL sum[maxn],a[maxn];
main() {
	LL n = read(), x = read();
	for(int i = 1;i <= n;++ i) a[i] = read(),sum[i] = sum[i - 1] + a[i],ans += 1ll * 5 * a[i] + 2ll * x;
	for(int k = 1;k < n;++ k) {
		LL tmp = k * x;
		for(int nxj,j = n,cnt = 1;j > 0;j = nxj,cnt ++) {
			nxj = std::max(j - k,0ll);
			LL s = sum[j] - sum[nxj];
			tmp += std::max(5ll,cnt * 2ll + 1ll) * s;
			if(tmp > ans) break;
		}
		ans = std::min(ans,tmp);
	}
	print(ans + 1ll * n * x); pc('\n');
	return 0;
}