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：
1 #include<cstdio>
2 #include<cstring>
3 #include<cmath>
4 #include<algorithm>
5 using namespace std;
6 const int MAXN = 100010;
7 int A[MAXN];
8 int main(){
9 int n,m;
10 while(~scanf("%d%d",&n,&m)){
11 for(int i = 0; i < n; i++){
12 scanf("%d",&A[i]);
13 }
14 sort(A, A + n);
15 int ans = 0x3f3f3f3f;
16 for(int i = n - m; i >= 0;i--){
17 int x = m - 1, temp = 0;
18 for(int j = i + m - 1; j >= i;j--){
19 temp += x * A[j];
20 // printf("x = %d a[%d] = %d temp = %d ",x,j,A[j],temp);
21 x -= 2;
22 }
23 // printf("i = %d\n",i);
24 ans = min(ans, temp);
25 }
26 printf("%d\n",ans);
27 }
28 return 0;
29 }