﻿PAT(A) 1104. Sum of Number Segments (20)-Ocrosoft

# 1104. Sum of Number Segments (20)

200 ms

65536 kB

16000 B

Standard

CAO, Peng

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 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
```

```#include <set>
#include <map>
#include <list>
#include <cmath>
#include <stack>
#include <queue>
#include <ctime>
#include <string>
#include <cstdio>
#include <vector>
#include <cctype>
#include <climits>
#include <sstream>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
template <typename T>
T GCD(T a, T b)
{
if (!b)return a;
return GCD(b, a%b);
}
template <typename T>
T LCM(T a, T b)
{
return a*b / GCD(a, b);
}

int main()
{
int n;cin>>n;
vector<double> v(n);
for(int i=0;i<n;i++)cin>>v[i];
double res=0.0;
for(int i=0;i<v.size();i++)
res+=(i+1)*(v.size()-i)*v[i]; // [(i-1)*(n-i+1)+1+(n-i)]*v[i]
printf("%.2lf\n",res);
return 0;
}```