1104. Sum of Number Segments (20) 找规律-优快云博客

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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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;
}