第五届华中区程序设计邀请赛暨武汉大学第十四届校赛 网络预选赛 A

Problem 1603 - Minimum Sum

Time Limit: 2000MS   Memory Limit: 65536KB  
Total Submit: 564  Accepted: 157  Special Judge: No

Description

There are n numbers A[1] , A[2] .... A[n], 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[n].( 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

 

题意： 长度为n的数组A 任意取 取出的数量为m 然后求   所有 abs( A[B[i]]-A[B[j]]  i<j ) 的和

题解： 比赛的时候 有一个思路  题目时间限制是2000ms 但woj 测评 2000+ms   过了

         先sort排序 取连续的长度为m的序列re[]   求其sum值 

     预处理求 re[1]-re[2] =ans[2]  并记录前缀和

                 re[1]-re[3] =ans[3]

                ...

                 re[1]-re[m]=ans[m]

    re[2]-re[3]=(re[1]-re[3])-(re[1]-re[2]) //

    re[2]-re[4]=(re[1]-re[4])-(re[1]-re[2]) //将要求的量 转化为已知量

 

 #include<iostream>
  #include<cstdio>
  #include<cmath>
  #include<algorithm>
  #include<cstring>
  using namespace std;
 int n,m;
 struct node
  {
      long long  w;
      int pos;
  }N[];
 long long ans[];
 long long  re[];
 long long  summ[];
 long long qq;
 long long exm;
 long long gg;
 int l=,r=;
 bool cmp1(struct node aa,struct node bb)
  {
      if(aa.w<bb.w)
      return true;
      return false;
  }
 long long fun( int ll,int rr)
  {
        l=ll;
        r=rr;
        int jishu=;
        for(int i=l;i<=r;i++)
            re[jishu++]=N[i].w;
            exm=;
            summ[]=;
            ans[]=;
          for(int i=;i<jishu;i++)
            {
              ans[i]=abs(re[]-re[i]);
              exm+=ans[i];
              summ[i]=exm;
            }
            qq=;
          for(int i=;i<jishu;i++)
          {
              qq=qq+summ[jishu-]-summ[i];
              qq=qq-ans[i]*(jishu-i-);
          }
      return qq;
  }
 int main()
  {
      while(scanf("%d %d",&n,&m)!=EOF)
      {
          memset(N,,sizeof(N));
          for(int i=;i<=n;i++)
          {
            scanf("%lld",&N[i].w);
            N[i].pos=i;
          }
          sort(N+,N++n,cmp1);
          gg=fun(,m);
          for(int i=;i<=n-m+;i++)
          {
              gg=min(gg,fun(i,i+m-));
          }
            printf("%lld\n",gg);
      }

  }