1104. Sum of Number Segments (20) 找规律

本文介绍了一种高效算法,用于计算给定正数数列中所有连续子序列(段)数值之和,并通过实例说明了如何避免O(n^2)的时间复杂度,确保算法在大数列输入下仍能快速得出结果。

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  1. Sum of Number Segments (20)

时间限制
200 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CAO, Peng
Given a sequence of positive numbers, a segment is defined to be a consecutive subsequence. For example, given the sequence {0.1, 0.2, 0.3, 0.4}, we have 10 segments: (0.1) (0.1, 0.2) (0.1, 0.2, 0.3) (0.1, 0.2, 0.3, 0.4) (0.2) (0.2, 0.3) (0.2, 0.3, 0.4) (0.3) (0.3, 0.4) (0.4).

Now given a sequence, you are supposed to find the sum of all the numbers in all the segments. For the previous example, the sum of all the 10 segments is 0.1 + 0.3 + 0.6 + 1.0 + 0.2 + 0.5 + 0.9 + 0.3 + 0.7 + 0.4 = 5.0.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N, the size of the sequence which is no more than 105. The next line contains N positive numbers in the sequence, each no more than 1.0, separated by a space.

Output Specification:

For each test case, print in one line the sum of all the numbers in all the segments, accurate up to 2 decimal places.

Sample Input:
4
0.1 0.2 0.3 0.4
Sample Output:
5.00

找规律的题,建立数组a[]存储数列,对数组中的每一个元素a[i]统计一下在i之前有多少,之后有多少就可以了。
如果使用O(n^2)的算法会超时。

#include<iostream>
#include<cstdio>
using namespace std;
const int maxn = 100001;
double a[maxn];
int main(){
    int n;
    double sum = 0;
    cin >> n;
    for(int i = 1;i <= n;i++)
        scanf("%lf",&a[i]);
    for(int i = 1;i <= n;i++){
        sum += a[i]*i*(n-i+1);   // put a[i] ahead to prevent from overflowing
    }
    printf("%.2f\n",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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