2014新生暑假个人排位赛09---470. diffsum

最新推荐文章于 2023-12-23 07:04:26 发布

原创最新推荐文章于 2023-12-23 07:04:26 发布 · 515 阅读

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本文介绍了一种算法问题——数组差值求和，通过排序和数学规律避免超时问题，给出了具体的C++实现代码。

时间限制 1000 ms 内存限制 65536 KB

题目描述

You are givin an array of integers, and you are to figure out the sum of differences between each pair of integers belonging to the array. SEE THE HINT FOR MORE INFORMATION.

输入格式

There are multiple test cases. The first line contains an integer n(n<=1e5), size of the array. The second line contains n integers, ai(|ai|<=100000), the array.

输出格式

For each case, output one line, the answer.

输入样例

4
1 1 2 2

输出样例

4

hint
for the test case the answer is abs(1-1)+abs(1-2)+abs(1-2)+abs(1-2)+abs(1-2)+abs(2-2)=4.

这道题不断地wa。结果还是因为int爆了，这样的问题已经被坑过很多次了，还是没有记住教训！然后是为了不超时不能暴力，这里又要用到数学规律了，多列几个式子，然后就是要考虑正负数以及绝对值，还要注意n=1的情况。

code：

#include
 <iostream>

#include
 <cstdio>

#include
 <cmath>

#include
 <algorithm>

usingnamespace

std;

longlong

a[100006];

intmain()

{        longlong 

n;

        while(~scanf("%lld",&n))

        {

            longlong

sum=0;

            if(n==1)

            {

                scanf("%lld",&a[0]);

                a[0]=abs(a[0]);

                printf("%lld\n",a[0]);

            }

            else

            {

                for(inti=0;i<n;i++)

                    scanf("%lld",&a[i]);

                sort(a,a+n);

                for(inti=0;i<n;i++)

                    sum=sum+(a[i]*(n-1-i)-a[i]*i);

                sum=abs(sum);

                printf("%lld\n",sum);

            }

        }

    return0;

}