1113 Integer Set Partition (25 分)

Given a set of N (>) positive integers, you are supposed to partition them into two disjoint sets A​1​​ and A​2​​ of n​1​​ and n​2​​ numbers, respectively. Let S​1​​ and S​2​​ denote the sums of all the numbers in A​1​​ and A​2​​, respectively. You are supposed to make the partition so that ∣ is minimized first, and then ∣ is maximized.

Input Specification:

Each input file contains one test case. For each case, the first line gives an integer N (2), 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 2​31​​.

Output Specification:

For each case, print in a line two numbers: ∣ and ∣, 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

 

 

#include<bits/stdc++.h>
using namespace std;
const int maxn=1010;
struct node{
    int n,s;
};
map<int,node> mp;
vector<int> v;
int getSumL(int index){
    int sum=0;
    for(int i=0;i<=index;i++){
        sum+=v[i];
    }
    return sum;
}
int getSumR(int index){
    int sum=0;
    for(int i=index;i<v.size();i++){
        sum+=v[i];
    }
    return sum;
}

int main(){
    int n;
    cin>>n;
    v.resize(n);
    for(int i=0;i<n;i++){
        cin>>v[i];
    }
    sort(v.begin(),v.end());
    int cnt;
    int sum1=0,sum2=0;
    if(n%2==0){
        cnt=0;
        for(int i=0;i<v.size();i++){
            if(i<n/2){
                sum1+=v[i];
            }
            else{
                sum2+=v[i];
            }
        }
    }
    else{
        cnt=1;
        for(int i=0;i<v.size();i++){
            if(i<n/2){
                sum1+=v[i];
            }
            else{
                sum2+=v[i];
            }
        }
    }
    cout<<cnt<<" "<<abs(sum1-sum2)<<endl;
    return 0;
}

一道水题，却被我想的复杂，真是无语