PAT (Advanced Level) Practise 1113 Integer Set Partition (25)

1113. Integer Set Partition (25)

時間限制

150 ms

記憶體限制

65536 kB

代碼長度限制

16000 B

判題程式

Standard

作者

CHEN, Yue

Given a set of N (> 1) positive integers, you are supposed to partition them into two disjoint sets A1 and A2 of n1 and n2numbers, respectively. Let S1 and S2 denote the sums of all the numbers in A1 and A2, respectively. You are supposed to make the partition so that |n1 - n2| is minimized first, and then |S1 - S2| is maximized.

Input Specification:

Each input file contains one test case. For each case, the first line gives an integer N (2 <= N <= 105), and then N positive integers follow in the next line, separated by spaces. It is guaranteed that all the integers and their sum are less than 231.

Output Specification:

For each case, print in a line two numbers: |n1 - n2| and |S1 - S2|, separated by exactly one space.

Sample Input 1:

10

23 8 10 99 46 2333 46 1 666 555

Sample Output 1:

0 3611

Sample Input 2:

13

110 79 218 69 3721 100 29 135 2 6 13 5188 85

Sample Output 2:

1 9359

排個序，偶數選是中間，奇數是中間靠前。

#include<cstdio>
#include<vector>
#include<queue>
#include<string>
#include<map>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long LL;
const int INF = 0x7FFFFFFF;
const int maxn = 1e5 + 10;
int n, a[maxn], sum, ans;

int main()
{
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++)
  {
    if (i <= n / 2 ) ans += a[i];
    sum += a[i];
  }
  if (n & 1) printf("1 %d\n", sum - ans * 2);
  else printf("0 %d\n", sum - ans * 2);
  return 0;
}