1104 Sum of Number Segments

本文分享了SumofNumberSegments题目的解题思路与过程,介绍了从暴力解法到优化算法的转变,揭示了每个数被加次数的规律,并通过代码实现展示了正确的计算顺序对结果的重要性。

题目: Sum of Number Segments

思路: 本来是不打算写这道题的博客的,但是解题过程有点曲折,所以还是写下博客记录下来。

第一次尝试了下暴力解法,时间复杂度为O(n2),理所当然地超时了,
很明显要寻找规律,画了图发现每个数被加的次数是有规律的,如果下标从零开始,对于下标为 i 的数,次数是 (i+1) * (n-i) ,然后累加就行。
但我发现后两个测试点始终过不了,然后我就去查网上其他大神的代码,检查了很多遍一直没发现有啥不同,最后一句一句对比提交,发现是乘的顺序不对👩。。。
形式上我的是 int * double ,别人的是 double * int, 可能是因为后一个精度更高?有大神解释一下么?
下面是代码
CODE ⇩⇩⇩

#include <bits/stdc++.h>
using namespace std;
int main() {
    int n;
    double c,sum{};
    scanf("%d",&n);
    for(int i = 0;i < n;i++) {
        scanf("%lf",&c);
        sum += c * (i+1) * (n-i); // 原来是(i+1) * (n-i) * c
    }
    printf("%.2f",sum);
    return 0;
}

Yousef has an array a of size n . He wants to partition the array into one or more contiguous segments such that each element ai belongs to exactly one segment. A partition is called cool if, for every segment bj , all elements in bj also appear in bj+1 (if it exists). That is, every element in a segment must also be present in the segment following it. For example, if a=[1,2,2,3,1,5] , a cool partition Yousef can make is b1=[1,2] , b2=[2,3,1,5] . This is a cool partition because every element in b1 (which are 1 and 2 ) also appears in b2 . In contrast, b1=[1,2,2] , b2=[3,1,5] is not a cool partition, since 2 appears in b1 but not in b2 . Note that after partitioning the array, you do not change the order of the segments. Also, note that if an element appears several times in some segment bj , it only needs to appear at least once in bj+1 . Your task is to help Yousef by finding the maximum number of segments that make a cool partition. Input The first line of the input contains integer t (1≤t≤104 ) — the number of test cases. The first line of each test case contains an integer n (1≤n≤2⋅105 ) — the size of the array. The second line of each test case contains n integers a1,a2,…,an (1≤ai≤n ) — the elements of the array. It is guaranteed that the sum of n over all test cases doesn't exceed 2⋅105 . Output For each test case, print one integer — the maximum number of segments that make a cool partition. Example InputCopy 8 6 1 2 2 3 1 5 8 1 2 1 3 2 1 3 2 5 5 4 3 2 1 10 5 8 7 5 8 5 7 8 10 9 3 1 2 2 9 3 3 1 4 3 2 4 1 2 6 4 5 4 5 6 4 8 1 2 1 2 1 2 1 2 OutputCopy 2 3 1 3 1 3 3 4 Note The first test case is explained in the statement. We can partition it into b1=[1,2] , b2=[2,3,1,5] . It can be shown there is no other partition with more segments. In the second test case, we can partition the array into b1=[1,2] , b2=[1,3,2] , b3=[1,3,2] . The maximum number of segments is 3 . In the third test case, the only partition we can make is b1=[5,4,3,2,1]
最新发布
06-09
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