PAT甲级1104

1104. Sum of Number Segments (20)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

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:

40.1 0.2 0.3 0.4 

Sample Output:

5.00

#include<iostream>#include<vector>#include<cstdio>#include<cstring>#include<string>#include<algorithm>#include<iomanip>using namespace std;/*double sum1(vector<double>a, int i, int j){vector<double> v;for (int k = i; k <= j; k++){if(k==i)v.push_back(a[k]);else{v.push_back(a[k]+v[k-i-1]);}}double sum = 0.0;for (int i = 0; i < v.size(); i++){sum += v[i];}return sum;}*/int main(){int N;cin >> N;double t;int n = N;vector<double> v;while (N--){cin >> t;v.push_back(t);}//sort(v.begin(), v.end(),cmp);不是排序//reverse(v.begin(), v.end());double sum = 0;/*for (int i = 0; i < v.size(); i++){sum += sum1(v, i, v.size() - 1);}*//*vector<double> vv;for (int i = 0; i < v.size(); i++){if (i == 0){vv.push_back(v[i]);}elsevv.push_back(v[i] * (i+1) + vv[i - 1]);估计有精度损失，其实这个跟下面的思路一样}*/for (int i = 0; i < v.size(); i++){sum += v[i]*(n-i)*(i+1);//找出元素出现的规律即可}printf("%.2f", sum);return 0;}