QBXT 二月五号整理

给你一列数, 询问和最大的子串. N<=10^6

 1 // N <=10^6
 2 #include<cstdio>
 3 #include<iostream>
 4 using namespace std;
 5 int n,a[105200];
 6 int maxSubstr(){
 7     int sum=0,minsum=0,answer=0;
 8     for(int i=1;i<=n;++i){
 9         sum+=a[i];
10         minsum=min(sum,minsum);
11         answer=max(answer,sum-minsum);
12     }
13     printf("%d\n",answer);
14 }
15 int main()
16 {
17     scanf("%d",&n);
18     for(int i=1;i<=n;i++)
19       scanf("%d",&a[i]);
20     maxSubstr();
21     return 0;
22 }

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.

 1 /*二、给你两列数 A; B, 定义一个子串 [l; r] 的权值为
 2    ∑r i=l Ai
 3 k= ___________
 4    ∑r i=l Bi
 5 问权值最大的子串.*/
 6 const double inf = 1e32;
 7 int n, a[maxn], b[maxn];
 8 double c[maxn];
 9 double maxSubstr() {
10     double sum = 0, minsum = inf, answer = -inf;
11     for (int i = 1; i <= n; ++ i) {
12         sum += c[i];
13         answer = max(answer, sum - minsum);
14         minsum = min(sum, minsum);
15     }
16     return answer;
17 }
18 double erfen() {
19     double l(-inf), r(inf);
20     while (l + 1e-10 < r) {
21         double mid = (l + r)/ 2;
22         for (int i = 1; i <= n; ++ i) {
23             c[i] = a[i] - mid * b[i];
24         }
25         if (maxSubstr() >= 0) {
26             l = mid;
27         } else {
28             r = mid - 1e-10;
29         }
30     } //把分母乘过去二分答案
31 }