Mike and gcd problem

    Mike has a sequence A = [a1, a2, ..., an] of length n. He considers the sequence B = [b1, b2, ..., bn] beautiful if the gcd of all its elements is bigger than 1, i.e. .

    Mike wants to change his sequence in order to make it beautiful. In one move he can choose an index i (1 ≤ i < n), delete numbers ai, ai + 1 and put numbers ai - ai + 1, ai + ai + 1 in their place instead, in this order. He wants perform as few operations as possible. Find the minimal number of operations to make sequence A beautiful if it's possible, or tell him that it is impossible to do so.

    is the biggest non-negative number d such that d divides bi for every i (1 ≤ i ≤ n).
Input

    The first line contains a single integer n (2 ≤ n ≤ 100 000) — length of sequence A.

    The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109) — elements of sequence A.
Output

    Output on the first line "YES" (without quotes) if it is possible to make sequence A beautiful by performing operations described above, and "NO" (without quotes) otherwise.

    If the answer was "YES", output the minimal number of moves needed to make sequence A beautiful.
Example
    Input

    2
    1 1

    Output

    YES
    1

    Input

    3
    6 2 4

    Output

    YES
    0

    Input

    2
    1 3

    Output

    YES
    1

Note

    In the first example you can simply make one move to obtain sequence [0, 2] with .

    In the second example the gcd of the sequence is already greater than 1.

题意： 假设小明有n场考试，已经考了k场，，剩下的几场考试  考试分数不能超过p分，n场考试的分数和不能超过x分   ，n场考试的中位数必须大于等于y分，，（n这里为奇数）

思路： 第一我们只需要找到已经考了的k场考试中是否小于y的个数已经>n/2，或者总分数是否已经超过x，，如果两者成立其一那么输出-1.
否则, 贪心，我们想要最后的分数越小越好，所以每次考试只考1分知道考试中小于y的个数==n/2  那么之后的考试都考y分，就可以保证中位数大于等于y  并且  总分数最小。

代码：

#include
int arr[1005];
void Print(int k,int n){
    int i;
    for(i=k;in/2||sum>x){ 
        printf("-1\n");
        return 0;
    }
    if(cnt==n/2){ //其余考试为y，则总成绩最小
       for(i=k;ix){
         printf("-1\n");
       }
       else Print(k,n);
    }
    else{
        for(i=k;ix) printf("-1\n");
        else Print(k,n);
    }
return 0;
}