1113 Integer Set Partition (25point(s)) - C語言 PAT 甲級

1113 Integer Set Partition (25point(s))

Given a set of N (>1) positive integers, you are supposed to partition them into two disjoint sets A₁ and A​₂ of n​₁​​ and n​₂​​ numbers, 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​​−S​₂​​∣ 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​​−S​₂​​∣, separated by exactly one space.

Sample Input:

10
23 8 10 99 46 2333 46 1 666 555

Sample Output:

0 3611

Sample Input:

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

Sample Output:

1 9359

題目大意：

輸入 N 個數，分爲兩個集合，使兩個集合元素個數相差最小，元素之和相差最大

輸出相差個數 ∣n₁​​−n₂​​∣，元素和的差 ∣S₁​​−S​₂​​∣

設計思路：

元素排序，前一半爲集合 1，後一半爲集合 2，計算輸出

編譯器：C (gcc)

#include <stdio.h>

int cmp(const void *a, const void *b)
{
        return *((int *)a) - *((int *)b);
}

int main(void)
{
        int n, a[100010];
        int i;
        int s1 = 0, s2 = 0;

        scanf("%d", &n);
        for (i = 0; i < n; i++)
                scanf("%d", a + i);
        qsort(a, n, sizeof(a[0]), cmp);
        for (i = 0; i < n / 2; i++)
                s1 += a[i];
        for (i; i < n; i++)
                s2 += a[i];
        printf("%d %d", n % 2, s2 - s1);
        return 0;
}