PAT Class 1104 Sum of Number Segments (20 points) (pit, int * int may overflow)

1104 Sum of Number Segments (20分)

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

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

Meaning of the questions:

Given a positive number series, which we can intercept any of several successive numbers, called clips. For example, a given number sequence {0.1, 0.2, 0.3, 0.4}, we have (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) The 10 fragments. Given positive integer sequence, and obtains the number of all fragments contain all. As the sum of the present embodiment 10 fragment is 0.1 + 0.3 + 0.6 + 1.0 + 0.2 + 0.5 + 0.9 + 0.3 + 0.7 + 0.4 = 5.0, the output of the sequence in a row all fragments contained and the number of the nearest decimal 2 bit ~

answer:

将数列中的每个数字读取到temp中，假设我们选取的片段中包括temp，且这个片段的首尾指针分别为p和q，那么对于p，有i种选择，即12…i，对于q，有n-i+1种选择，即i, i+1, … n，所以p和q组合形成的首尾片段有i * (n-i+1)种，因为每个里面都会出现temp，所以temp引起的总和为temp * i * (n – i + 1)；遍历完所有数字，将每个temp引起的总和都累加到sum中，最后输出sum的值～
————————————————
版权声明：本文为CSDN博主「柳婼」的原创文章，遵循 CC 4.0 BY-SA 版权协议，转载请附上原文出处链接及本声明。
原文链接：https://blog.csdn.net/liuchuo/article/details/51985824

真的很细节挖了好多次。 int*int*double 转换是 int*int 变为 int ,然后 int *double

但是当 i 去100000时 int *int 可能会溢出，然后就答案错误，解决方法要么把int换为long long 要么把 double 提前

使得 double * int 变为 double
————————————————
版权声明：本文为CSDN博主「Cute_jinx」的原创文章，遵循 CC 4.0 BY-SA 版权协议，转载请附上原文出处链接及本声明。
原文链接：https://blog.csdn.net/Cute_jinx/article/details/82497800

AC代码：

#include<bits/stdc++.h>
using namespace std;
double a;
int n;
double s=0.0;
int main(){
    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>a;
        s+=a*i*(n-i+1);    
        //int*int*double 转换是 int*int 变为 int ,然后 int *double
        //当 i 去100000时 int *int 可能会溢出，然后就答案错误，
        //解决方法要么把int换为long long 要么把 double 提前 使得 double * int 变为 double 
        //s+=(n-i+1)*i*a最后两个测试点过不了    
    }
    printf("%.2f",s);
    return 0;
}