pat 1104. Sum of Number Segments

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 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

**思路:找规律的题，第i个值出现次数为(i+1)乘(n-i)，累加即可

PS:记得去年做的时候找了半天都没看出规律，今年几分钟就看出来了,

写代码还是需要勤加练习。**

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn=100005;
double a[maxn],sum=0;
int n;
int main(){
//freopen("in.txt","r",stdin);
scanf("%d",&n);
for(int i=0;i<n;i++){
scanf("%lf",&a[i]);
sum+=a[i]*(n-i)*(i+1);
}
printf("%.2f\n",sum);
return 0;
}