【PAT】A. Sum of Number Segments (20)

最新推荐文章于 2021-05-06 20:36:14 发布

原创最新推荐文章于 2021-05-06 20:36:14 发布 · 771 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#PAT

本文介绍了一种高效计算数列所有连续子序列（段）数值之和的方法。通过分析每个数字在不同子序列中出现的频次规律，避免了暴力求解带来的超时问题。给出的C++代码实现了该算法，并通过样例验证了其正确性和精度。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

A. Sum of Number Segments (20)
时间限制
200 ms
内存限制
65536 kB
代码长度限制
8000 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

解题思路：
1，暴力的方法会超时。计算每个数字出现的次数，将公式推导出来即可,比如总共有4个数字，分别出现为

i	1	2	3	4
0	n-0	n-1	n-2	n-3
1	-	n-1	n-2	n-3
2	-	-	n -2	n-3
3	-	-	-	n-3

则第i个数字加和的个数为

(i + 1) * (n - i)

$(i+1)*(n-i)$

所以程序如下

#include <iostream>
using namespace std;
int main(){

    int n;
    scanf("%d", &n);

    double a;
    double counter = 0;
    double sum = 0;

    for (int i = 0; i < n; i++){
        scanf("%lf", &a);
        counter = (n - i);  //这个地方不能用int，防止整型溢出
        counter *= (i + 1);
        sum += counter * a;
    }

    printf("%.2lf", sum);


}