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

<c++> #include<stdio.h> #include<string.h> #include<iostream> #include<algorithm> 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<n; i++) cin >> a[i]; int gc = a[0]; for(int i=1; i<n; i++){ gc = gcd(gc,a[i]); } if(gc > 1){ printf("YES\n0\n"); } else{ int res = 0; for(int i=0; i<n-1; 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; i<n; i++){ if(a[i]%2 == 1){ res += 2; } } printf("YES\n"); printf("%d\n",res); } return 0; } </c++>