Day1-B-CF-1144B

简述：有一个n个元素的序列，选奇数下一个就选偶数，偶数则下一个就是奇数，问能否取完，能取完输出0，否则输出能剩下的最小的之和

思路：统计奇偶数个数，若相等或相差一则取完，否则排列后取出最小的前x个(x为相差个数，x > 1)

代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<vector>

using namespace std;

const int maxm = 2019;
int buff[maxm];

int main() {
    int n;
    scanf("%d",&n);
    for(int i = 0;i < n;++i) {
        scanf("%d",&buff[i]);
    }
    vector<int> even,odd;
    for(int i = 0;i < n;++i) {
        int tmp = buff[i];
        if(tmp % 2 == 0)
            even.push_back(tmp);
        else
            odd.push_back(tmp);
    }
    int evenlength = even.size(),oddlength = odd.size();
    if(evenlength == oddlength)
        printf("0");
    else if(evenlength == oddlength - 1 || evenlength == oddlength + 1) {
        printf("0");
    }
    else {
        int length = abs(evenlength - oddlength);
        sort(even.begin(),even.end()), sort(odd.begin(), odd.end());
        if(evenlength > oddlength) {
            int val1 = 0,val2 = 0;
            for(int i = 0;i < length ;++i)
                val1 += even[i];
            for(int i = 0;i < length-1 ;++i)
                val2 += even[i];
            val1 = min(val1,val2);
            printf("%d",val1);
        } else {
            int val1 = 0,val2 = 0;
            for(int i = 0;i < length ;++i)
                val1 += odd[i];
            for(int i = 0;i < length-1 ;++i)
                val2 += odd[i];
            val1 = min(val1,val2);
            printf("%d",val1);
        }
    }
    return 0;
}