PAT甲级1104(子列之和)

1104 Sum of Number Segments (20 point(s))

  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 10^​5. 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

题目大意

  给定一个数列，求其所有连续子列之和，结果精确到两位小数

解题思路

  很简单的一道题，多写几个例子找规律，很容易就知道每个数字最终会被加(i + 1) * (N - i)次

代码

#define _CRT_SECURE_NO_WARNINGS
#include 

int main()
{
	int N, i;
	double num, sum = 0;
	int ret = scanf("%d", &N);
	for (i = 0; i < N; ++i) {
		ret = scanf("%lf", &num);
		sum += num * (i + 1) * (N - i);
	}
	printf("%.2f", sum);
	return 0;
}