PAT-A 真题 – 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) 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

题目大意：给定n个数字，求这n个数字所有相邻序列的和。

直接暴力会超时，正确的做法是分析一下在所有序列中，每个数字被加了几次。

当序列元素个数为1时，第一个元素加了1*1次

当序列元素个数为2时，第一个加了1*2次，第二个加了2*1次

当序列元素个数为3时，第一个加了1*3次，第二个加了2*2次，第三个加了3*1次

通过这样的规律，我们就能写出以下程序：

#include <cstdio>
#include <iostream>
using namespace std;

int main(){
  int n;
  cin >> n;
  double ans = 0.0;
  for(int i = 1, x = n; i <= n; i++, x--){
    double input, co = 1.0 * i * x;  //不加1.0会导致后两个测试点error 
    cin >> input;
    ans += co * input;
  }
  printf("%.2f", ans);
  return 0;
}