三分
题目:
Weakness and Poorness
题意:
给一串整数 a1,a2,...,an ,求在所有 a1−x,a2−x,...,an−x 串中的最小Weakness值,Weakness定义为是对该串所有子串,求子串中数字的和的绝对值,其中的最大值
思路:
- 所有子串的和中,求最大值,这个是经典问题,O(n)可以解决,扫一遍序列,将当前值加入累计值,如果累计值为负就对累计值置零;
- 这里题目问的是子串的和的绝对值的最大值,这个也不难,对每个数取相反数,再做一次上述步骤,取两次结果的较大值即可
- 问题的难点在于,如何找到x,稍微计算一下样例,可以发现,串的Weakness值是先随着x的增大而减小,到达某个临界值后随着x的增大而增大,一个谷函数,三分枚举,O(n)验证,完成
代码:
#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-8;
const int maxn = 2e5 + 5;int arr[maxn];
double arr1[maxn];
int n;double calc(double arr[]){double t = 0;double res = arr[0];for (int i=0;i<n;++i){t = arr[i] + t;res = max(res, t);if (t < 0+eps){t = 0;}}return res;
}double solve(double t){for (int i=0;i<n;++i)arr1[i] = arr[i] - t;double x1 = calc(arr1);for (int i=0;i<n;++i)arr1[i] = -arr1[i];double x2 = calc(arr1);double x = max(fabs(x1), fabs(x2));return x;
}
int main(){scanf("%d",&n);int maximum = -10001, minimum = 10001;for (int i=0;i<n;++i){scanf("%d",arr+i);maximum = max(arr[i], maximum);minimum = min(arr[i], minimum);}int dcnt = 100;double l = minimum, r = maximum;while (dcnt--){double m1 = l + (r-l)/3, m2 = l + (r-l)*2/3;double x1 = solve(m1), x2 = solve(m2);if (x1 > x2){l = m1;}else r = m2;}double res=solve(l);printf("%.8f\n",res);return 0;}
更多推荐
三分
发布评论