Tenka 1 Computer Contest C-Align

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C - Align

Time limit : 2sec / Memory limit : 1024MB

Score : 400 points

Problem Statement

You are given N integers; the i-th of them is Ai. Find the maximum possible sum of the absolute differences between the adjacent elements after arranging these integers in a row in any order you like.

Constraints

• 2≤N≤105
• 1≤Ai≤109
• All values in input are integers.

Input

Input is given from Standard Input in the following format:

NA1:AN

Output

Print the maximum possible sum of the absolute differences between the adjacent elements after arranging the given integers in a row in any order you like.

Sample Input 1Copy

568123

Sample Output 1Copy

21

When the integers are arranged as 3,8,1,6,2, the sum of the absolute differences between the adjacent elements is |3?8|+|8?1|+|1?6|+|6?2|=21. This is the maximum possible sum.

Sample Input 2Copy

6314159

Sample Output 2Copy

25

Sample Input 3Copy

3551

Sample Output 3Copy

8

題解：分奇數和偶數討論。當為奇數時，一定是中間的兩個數在左右邊（兩種情況）使得結果最大。偶數同理，更好枚舉，只有一種情況。

#include <iostream>#include <cstring>#include <algorithm>#include <cmath>#include <cstdio>#include <vector>#include <queue>#include <set>#include <map>#include <stack>#define ll long long//#define localusing namespace std;const int MOD = 1e9+7;const int inf = 0x3f3f3f3f;const double PI = acos(-1.0);const int maxn = 1e5+10;int main() {#ifdef local    if(freopen("/Users/Andrew/Desktop/data.txt", "r", stdin) == NULL) printf("can‘t open this file!\n");#endif        int n;    int a[maxn];    ll mx = 0;    ll sum[maxn];    scanf("%d", &n);    for (int i = 0; i < n; ++i) {        scanf("%d", a+i);    }    sort(a, a+n);    for (int i = 0; i < n; ++i) {        if (!i) sum[i] = a[i];        if (i) sum[i] = sum[i-1]+a[i];    }    if (n&1) {        if (n == 3) {            mx = max(abs(2*a[2]-a[1]-a[0]), abs(2*a[0]-a[1]-a[2])); //當n=3時，特殊討論一下        } else {            //如果是選擇靠左邊的兩個數作為兩個端點，那麼他們一定是小於它們相鄰的數的。            //同理，如果是選擇靠左邊的兩個數作為兩個端點，那麼他們一定是大於它們相鄰的數的。            //將需要算兩次的數*2            mx = max(abs(2*(sum[n-1]-sum[n/2])-2*(sum[n/2-2])-(a[n/2]+a[n/2-1])), abs(2*(sum[n-1]-sum[n/2+1])-2*(sum[n/2-1])+(a[n/2]+a[n/2+1])));        }    } else {        mx = abs(2*(sum[n-1]-sum[n/2])-2*(sum[n/2-2])+abs(a[n/2]-a[n/2-1]));    }    printf("%lld\n", mx);#ifdef local    fclose(stdin);#endif    return 0;}

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Tenka 1 Computer Contest C-Align