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 1. 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

题意：

　　给出一组数字，求这组数字各个连续子段的和。

思路：

　　由数组中的一个数字组成的片段假设是temp：1，2，3，4，5，temp（i），7，8，9。则该片段的起始位置有 i 种不同的可能，结束位置有（n - i + 1）种不同的可能。所以，temp出现的总的次数就是 i * (n - i + 1)。遍历数组然后将不同的temp出现的次数相加即可。

Code：

#include <bits/stdc++.h>

using namespace std;

int main() {
    int n;
    cin >> n;
    double ans = 0.0, temp;

    for (int i = 1; i <= n; ++i) {
        cin >> temp;
        ans = ans + temp * i * (n - i + 1);
    }

    cout << fixed << setprecision(2) << ans << endl;
    return 0;
}

　　刚开始想着用遍历的方法来做，但是这样做的话最后两组数据不能够通过。后来想到用数学的方法，但是自己想到的方法太弱，还是没能过。后来就看别人的代码了，现在觉得自己还是那么菜。

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2020-07-12 21:41:57

　　刚才又提交了一下这道题，发现上次全部通过的测试点，这次有一组数据没有通过。后来发现原来是被一位大佬找到了错误修改了测试点，大佬的水平我只能够仰望。https://blog.zhengrh.com/post/about-double/

AC代码：

#include <bits/stdc++.h>

using namespace std;

int main() {
    int n;
    cin >> n;
    long long ans = 0;
    double temp;

    for (int i = 1; i <= n; ++i) {
        cin >> temp;
        ans += (long long)(temp * 1000) * i * (n - i + 1);
    }

    cout << fixed << setprecision(2) << (double)(ans / 1000.0) << endl;
    return 0;
}

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