Minimum Sum（思维）

Problem 1603 - Minimum Sum

Time Limit: 2000MS   Memory Limit: 65536KB    Total Submit: 563  Accepted: 156  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

题解：题意就是求连续m个数字相互差的绝对值最小；

刚开始就想着先排序，从大到小模拟了下找到了3x1+x2-x3-3x4;由此可以看出规律；但是由于想着数据是1e5，就不敢写。。。然后队友写了下就过了。。。吊。。。其实是因为当N是1e5的时候，如果M也很大，那么i是从n-m开始的，所以也没啥事；

然后的思路就是从后往前大暴力。。。

代码：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int MAXN = 100010;
int A[MAXN];
int main(){
    int n,m;
    while(~scanf("%d%d",&n,&m)){
        for(int i = 0; i < n; i++){
            scanf("%d",&A[i]);
        }
        sort(A, A + n);
        int ans = 0x3f3f3f3f;
        for(int i = n - m; i >= 0;i--){
            int x = m - 1, temp = 0;
            for(int j = i + m - 1; j >= i;j--){
                temp += x * A[j];
            //    printf("x = %d a[%d] = %d temp = %d ",x,j,A[j],temp);
                x -= 2;
            }
        //    printf("i = %d
",i);
            ans = min(ans, temp);
        }
        printf("%d
",ans);
    }
    return 0;
}