Polycarp has an array aa consisting of nn integers. He wants to play a game with this array. The game consists of several moves. On the first move he chooses any element and deletes it (after the first move the array contains n−1n−1 elements). For each of the next moves he chooses any element with the only restriction: its parity should differ from the parity of the element deleted on the previous move. In other words, he alternates parities (even-odd-even-odd-… or odd-even-odd-even-…) of the removed elements. Polycarp stops if he can’t make a move. Formally:
If it is the first move, he chooses any element and deletes it; If it is the second or any next move:
if the last deleted element was odd, Polycarp chooses any even element and deletes it; if the last deleted element was even, Polycarp chooses any odd element and deletes it.If after some move Polycarp cannot make a move, the game ends.
Polycarp’s goal is to minimize the sum of non-deleted elements of the array after end of the game. If Polycarp can delete the whole array, then the sum of non-deleted elements is zero. Help Polycarp find this value. Input The first line of the input contains one integer nn (1≤n≤20001≤n≤2000) — the number of elements of aa. The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (0≤ai≤1060≤ai≤106), where aiai is the ii-th element of aa.
Output Print one integer — the minimum possible sum of non-deleted elements of the array after end of the game. ExamplesInput
5 1 5 7 8 2
Output
0
Input
6 5 1 2 4 6 3
Output
0
Input
2 1000000 1000000
Output
1000000
此题意思就是奇偶数相互抵消，最后求出剩余数的最小和。所以为了得到最小和，那我们在奇偶数抵消的过程中就要找大的数相互抵消，一种情况是如果奇数的个数和偶数的个数相等或只相差一，那么最后所有数都会被抵消掉，那么和就为零了，就不需要考虑其中数大小的关系了；另一种情况就是看是奇数个数多还是偶数的个数多，为了得到最小的和，那就把个数多的那个从小到大排序，并将最后消除后会剩余的个数求出来，并相加那么多个，就可以啦。
#include"stdio.h" #include"algorithm" using namespace std; int main() { int n; int a[2005]; while(~scanf("%d",&n)) { int sum1=0; int sum2=0; for(int i=0;i < n;i ++) { scanf("%d",&a[i]); } for(int i=0;i < n;i ++) { if(a[i]%2==0) { sum1 ++; } else { sum2 ++; } } int m=sum1-sum2; int l=sum2-sum1; int ss=0; int c=0; if(sum1==sum2||m==1||l==1) { printf("0\n"); } else { sort(a,a+n); if(m > 0) { m--; for(int i = 0;i < n;i ++) { if(a[i] % 2 == 0) { ss+=a[i]; m--; } if(m==0) { break; } } printf("%d\n",ss); } else if(l > 0) { l--; for(int i = 0;i < n;i ++） { if(a[i] % 2 != 0) { c+=a[i]; l--; } if(l==0) { break; } } printf("%d\n",c); } } } return 0; }