Mike and gcd problem Codeforces Round #410 (Div. 2)

题目原文

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.
Examples
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.

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题目大意：

对一个序列A：{a1,a2,a3,…ai,ai+1….an}，通过操作{ai,ai+1} -> {ai - ai+1, ai + ai+1}
要求最后gcd{A}>1，求最小操作数量。

题目分析：

要想使gcd>1，必然是一个全偶数的序列。题目就变成了把一个序列转变成全偶数数列。

设a,b；
第一次操作后：a - b, a + b;
第二次操作后：-2b , 2a;
所以可以知道，如果是两个奇数，那么可以通过一次操作变成全偶数，一奇一偶则是两次。

所以最小操作数就是：先把所有的两个相邻奇数变成偶数，每次操作数+1，然后把剩余的奇数变成偶数，每次操作数+2。

AC代码

#include
#include
#include
#include
using namespace std;
int gcd(int x,int y)
{
    return y==0?x:gcd(y,x%y);
}
int a[1500000];
int main()
{
    int n;
    cin >> n;
    for(int i=0; i> a[i];
    int gc = a[0];
    for(int i=1; iif(gc > 1){
        printf("YES\n0\n");
    }
    else{
        int res = 0;
        for(int i=0; i1; i++){
            if(a[i]%2 == 1 && a[i+1]%2 == 1){
                res ++;
                int b = a[i];
                a[i] = -2 * a[i+1];
                a[i+1] = 2 * b;
            }
        }
        for(int i=0; iif(a[i]%2 == 1){
                res += 2;
            }
        }
        printf("YES\n");
        printf("%d\n",res);
    }

    return 0;
}
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