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 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
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
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
#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;
}