PAT 甲级 A1104

1104 Sum of Number Segments (20分)

题目描述

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.

输入格式

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 $10^{5}$. The next line contains N positive numbers in the sequence, each no more than 1.0, separated by a space.

输出格式

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

总结

1. 这并不是一道排列组合找区间的题，实际上，这是一道让你找规律的题。
2. 可以按照以下规律，找出数组长度为n时的规律。不难发现，下标为i的元素出现的次数为(i+1)*(n-i)，然后再将次数乘元素值，累加可得到结果
3. 但有一个问题，如果先用一个变量保存出现次数，再乘出现的次数，测试点3，和4过不去，如下代码注释所示。问题还望大佬帮我解决。

元素序号	1	2	3	4	5
片段长度为1时出现次数	1	1	1	1	1
片段长度为2时出现次数	1	2	2	2	1
片段长度为3时出现次数	1	2	3	2	1
片段长度为4时出现次数	1	2	2	2	1
片段长度为5时出现次数	1	1	1	1	1
总次数	5	8	9	8	5

AC代码

#include<iostream>
using namespace std;
int main() {
	int n;
	double ans = 0.0,temp;
	scanf("%d",&n);
	for (int i = 0; i < n; i++) {
		scanf("%lf", &temp);
		//long long cnt = (i + 1)*(n - i);   int可能越界,因此我分别试过long long和double
		//ans += temp*cnt;    若先计算出现次数再相乘测试点3,4无法通过
		ans += temp * (i + 1)*(n - i); //temp一定要写在前面,否则很可能越界
	}
    printf("%.2lf\n",ans);
	return 0;
}