Shopping
Time Limit: 1000MS Memory limit: 65536K
题目描述
Saya and Kudo go shopping together.
You can assume the street as a straight line, while the shops are some points on the line.
They park their car at the leftmost shop, visit all the shops from left to right, and go back to their car.
Your task is to calculate the length of their route.
You can assume the street as a straight line, while the shops are some points on the line.
They park their car at the leftmost shop, visit all the shops from left to right, and go back to their car.
Your task is to calculate the length of their route.
输入
The input consists of several test cases.
The first line of input in each test case contains one integer N (0<N<100001), represents the number of shops.
The next line contains N integers, describing the situation of the shops. You can assume that the situations of the shops are non-negative integer and smaller than 2^30.
The last case is followed by a line containing one zero.
The first line of input in each test case contains one integer N (0<N<100001), represents the number of shops.
The next line contains N integers, describing the situation of the shops. You can assume that the situations of the shops are non-negative integer and smaller than 2^30.
The last case is followed by a line containing one zero.
输出
For each test case, print the length of
their shopping route.
示例输入
4 24 13 89 37 6 7 30 41 14 39 42 0
示例输出
152 70
提示
Explanation for the first sample: They park
their car at shop 13; go to shop 24, 37 and 89 and finally return to shop 13. The total length is (24-13) + (37-24) + (89-37) + (89-13) = 152
来源
2010年山东省第一届ACM大学生程序设计竞赛
链接：http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2154
水题 排序
#include <iostream>
#include <cstring>
#include <cstdio>
#include <string>
#include <algorithm>
#include <cmath>
#define MAX 100005
#define LL long long
using namespace std;
int main()
{
int n;
int a[MAX];
while(~scanf("%d",&n)&&n!=0){
for(int i=0;i<n;i++){
scanf("%d",&a[i]);
}
sort(a,a+n);
LL sum=0;
for(int i=1;i<n;i++)
sum+=(a[i]-a[i-1]);
sum+=(a[n-1]-a[0]);
printf("%lld
",sum);
}
return 0;
}