UOJ #21. 缩进优化

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题意

每行a[i]个空格，找一个合适的Tab长度x，将每x个空格替换为Tab，问最小字节数

题解

大暴力

前缀和，记录x * (j + 1) - 1 ~ x * j的个数，再乘j

最后乘(x - 1)，统计ans

调试记录

找空格要Max + 100000，否则会漏解

#include <cstdio>
#include <algorithm>
#define maxn 2000005
#define int long long

using namespace std;

int n, a[maxn], Max = 0, sum = 0;

signed main(){
	scanf("%lld", &n);
	for (int x, i = 1; i <= n; i++) scanf("%lld", &x), a[x]++, Max = max(Max, x), sum += x;
	for (int i = 1; i <= maxn; i++) a[i] += a[i - 1];
	
	int ans = sum;
	for (int i = 1; i <= Max; i++){
		int tot = 0;
		for (int j = 1; j * i <= Max + 100000; j++)
			tot += (a[j * i - 1] - a[(j - 1) * i - 1]) * (j - 1);
		tot *= (i - 1);
		ans = min(ans, sum - tot);
	}
	printf("%lld\n", ans);
	return 0;
}