QBXT 二月五号整理

给你一列数, 询问和最大的子串. N<=10^6

// N <=10^6
#include<cstdio>
#include<iostream>
using namespace std;
int n,a[105200];
int maxSubstr(){
    int sum=0,minsum=0,answer=0;
    for(int i=1;i<=n;++i){
        sum+=a[i];
        minsum=min(sum,minsum);
        answer=max(answer,sum-minsum);
    }
    printf("%d\n",answer);
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
      scanf("%d",&a[i]);
    maxSubstr();
    return 0;
}

啊！！！好巧妙，幸亏当时记了笔记。。。

给你两列数 A; B, 定义一个子串 [l; r] 的权值为

∑r i=l Ai除以∑r i=l Bi 问权值最大的子串
n ≤ 10^5.

/*二、给你两列数 A; B, 定义一个子串 [l; r] 的权值为
   ∑r i=l Ai
k= ___________
   ∑r i=l Bi
问权值最大的子串.*/
const double inf = 1e32;
int n, a[maxn], b[maxn];
double c[maxn];
double maxSubstr() {
    double sum = 0, minsum = inf, answer = -inf;
    for (int i = 1; i <= n; ++ i) {
        sum += c[i];
        answer = max(answer, sum - minsum);
        minsum = min(sum, minsum);
    }
    return answer;
}
double erfen() {
    double l(-inf), r(inf);
    while (l + 1e-10 < r) { 
        double mid = (l + r)/ 2;
        for (int i = 1; i <= n; ++ i) {
            c[i] = a[i] - mid * b[i];
        }
        if (maxSubstr() >= 0) {
            l = mid;
        } else {
            r = mid - 1e-10;
        }
    } //把分母乘过去二分答案
}