1113 Integer Set Partition （sort排序）

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最新推荐文章于 2022-11-15 17:07:48 发布

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整数划分绝对值差排序算法集合优化动态分配

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1113 Integer Set Partition (25 分)

Given a set of N (>1) positive integers, you are supposed to partition them into two disjoint sets A1 and A2 of n1 and n2 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−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:

<span style="color:#404040"><span style="background-color:#ffffff"><code class="language-in">10
23 8 10 99 46 2333 46 1 666 555
</code></span></span>

Sample Output 1:

<span style="color:#404040"><span style="background-color:#ffffff"><code class="language-out">0 3611
</code></span></span>

Sample Input 2:

<span style="color:#404040"><span style="background-color:#ffffff"><code class="language-in">13
110 79 218 69 3721 100 29 135 2 6 13 5188 85
</code></span></span>

Sample Output 2:

<span style="color:#404040"><span style="background-color:#ffffff"><code class="language-out">1 9359</code></span></span>

题意：给定一系列数，把这些数分成两个不相交的集合s1，s2；要求s1，s2长度差的绝对值最小，数列和的差的绝对值最大。

思路：直接把输入的数组排序，那么s1就是前n/2个，s2就是剩下的，然后分别算和就好了。

#include <iostream>
#include "algorithm"
using namespace std;
long long int nums[100100];
int main() {
    int n;
    cin>>n;
    for (int i = 0; i < n; ++i) {
        cin>>nums[i];
    }
    int len1,len2,sum1=0,sum2=0;
    if (n%2==0) len1=len2=n/2;
    else len1=n/2+1,len2=n/2;
    sort(nums,nums+n);
    for (int i = 0; i < len2; ++i) {
        sum2 += nums[i];
    }
    for (int i = len2; i <len2+len1 ; ++i) {
        sum1 +=nums[i];
    }
    cout<<len1-len2<<" "<<sum1-sum2;
}