【Codeforces 798 C】+ gcd

C. Mike and gcd problem
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
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.

a ,b ;
a - b,a + b:
-2b,2a.

最多操作两次，最大公约数便会为 2

对于gcd(a,b) = 1时，必须操作 ，且操作后 gcd为 2

AC代码：

#include<cstdio>#include<cmath>using namespace std;typedef long long LL;LL a[100010];LL gcd(LL a,LL b){ return a % b ? gcd(b,a % b) : b; }int main(){    LL n,t;    scanf("%lld",&n);    for(int i = 1; i <= n; i++)        scanf("%lld",&a[i]);    LL ans = gcd(abs(a[1]),abs(a[2]));    for(int i = 3; i <= n; i++)        ans = gcd(ans,abs(a[i]));    if(ans > 1) printf("YES\n0\n");    else{        int num = 0;        for(int i = 1; i < n; i++)            while(a[i] & 1)                t = a[i],a[i] -= a[i + 1],a[i + 1] += t,++num;        while(a[n] & 1)            t = a[n - 1],a[n - 1] -= a[n],a[n] += t,++num;       printf("YES\n%d\n",num);    }    return 0;}