PAT甲级1104 Sum of Number Segments (20 分)-优快云博客

本文详细解析了1104 Sum of Number Segments问题的解题思路，通过分析序列中每个元素在不同片段中出现的次数，提供了一种高效的求解所有片段和的方法。

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1104 Sum of Number Segments (20 分)

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) and (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 Npositive 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:

0.1 0.2 0.3 0.4

Sample Output:

5.00

题目大意：

给出长度为n的数字序列，求所有片段的和。

思路：

将每个片段只考虑首尾,当首为i时有1 2 3 ``` I 共i种, 而尾有 I I + 1 ``` n 共n – I + 1 种可能.每次就枚举第i位的数,那么他共会出现I * (n – I + 1)次.

参考代码：

#include<cstdio>
int n;
double sum, temp;
int main(){
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i){
		scanf("%lf", &temp);
		sum += temp * i * (n - i + 1);
	}
	printf("%.2f", sum);
	return 0;
}

代码参考:https://blog.youkuaiyun.com/liuchuo/article/details/51985824