PAT甲级真题 1104 Sum of Number Segments (20分) C++实现（智力题，测试点3double精度坑点）

本文链接：https://blog.youkuaiyun.com/zhang35/article/details/107696051

本文介绍了一种计算序列中所有连续子序列（段）数值总和的算法，通过计算每个数字在所有段中出现的次数，实现高效求解。文章讨论了浮点数精度问题，并提供了两种解决方案：一是将输入值放大1000倍转化为整数计算；二是使用long double类型存储累加结果。

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题目

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

思路

智力题，可以直接算出每个数字出现的次数：(n - i) * (i + 1)次（从该数字开始向后有 n-i 种可能，该数字前面还有 i 个数可以和它组合），直接一次遍历即可得出结果。

坑点

测试点3报错，花了很久也没找出问题，看了网上其他同学答案，原来是浮点数精度问题，真的坑…

因为double存储十进制小数时是有些许误差的，当N很大也就是累加次数足够大时，这个误差就会体现出来导致结果出错。

柳神的方法是将输入值扩大1000倍后作为整型相加；也可以直接使用long double类型存储sum值。

代码

柳神方法，自行调整精度：

#include <iostream>
using namespace std;

int main() {
    int n;
    scanf("%d", &n);
    long long sum = 0;
    for (int i=0; i<n; i++){
        double temp;
        scanf("%lf", &temp);
        sum += (long long)(temp * 1000) * (n-i) * (i+1);
    }
    printf("%.2f", sum / 1000.0);
    return 0;
}

直接使用long double类型：

#include <iostream>
using namespace std;

int main() {
    int n;
    scanf("%d", &n);
    long double sum = 0.0;
    for (int i=0; i<n; i++){
        double temp;
        scanf("%lf", &temp);
        sum += temp * (n-i) * (i+1);
    }
    printf("%.2Lf", sum);
    return 0;
}