WHOJ　1603 - Minimum Sum【思维】

Problem 1603 - Minimum Sum

Time Limit: 2000MS   Memory Limit: 65536KB   
Total Submit: 561  Accepted: 154  Special Judge: No

Description

There are n numbers A[1] , A[2] .... A
, you can select m numbers of it A[B[1]] , A[B[2]] ... A[B[m]]  ( 1 <= B[1] < B[2] .... B[m] <= n ) such that Sum as small as possible.

Sum is sum of abs( A[B[i]]-A[B[j]] ) when 1 <= i < j <= m.

Input

There are multiple test cases.

First line of each case contains two integers n and m.( 1 <= m <= n <= 100000 )

Next line contains n integers A[1] , A[2] .... A
.( 0 <= A[i] <= 100000 )

It's guaranteed that the sum of n is not larger than 1000000.

Output

For each test case, output minimum Sum in a line.

Sample Input

4 2

5 1 7 10

5 3

1 8 6 3 10

Sample Output

2

8

AC-code：

<pre name="code" class="cpp">#include<cstdio>
#include<algorithm>
using namespace std;
int s[100005];
int main()
{
int n,m,i,k,j;
long long sum,ans;
while(~scanf("%d%d",&n,&m))
{
for(i=1;i<=n;i++)
scanf("%d",&s[i]);
sort(s+1,s+n+1);
ans=0x3f3f3f3f;
for(i=1;i<=n-m+1;i++)//每个可能的数列的起始位置
{
sum=0;
for(j=1;j<=m;j++)
sum+=(2*j-m-1)*s[i+j-1];//2*j-m-1=(j-1)-(m-j)即该数在该序列中前面的数的个数减去后面的数的个数
if(sum<ans)
ans=sum;
}
printf("%lld\n",ans);
}
return 0;
}