CodeForces
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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.
首先题干那种操作只能保证两个奇数出来一定是偶数
单个奇数变偶数只要两次没错就只能保证这个….
其他的啥都保证不了..除了先天出现的gcd还是别想了….
想到这个就暗全变偶数猜一发能过就过过不了算了
#include<iostream>#include<string>#include<algorithm>#include<cstdio>using namespace std;int tu[100001];int n;typedef long long ll;int gcd(int a, int b){ if (b == 0) return a; return gcd(b, a%b);}int main(){ cin >> n; for(int a=1;a<=n;a++)scanf("%d",&tu[a]); int gg = tu[1]; for (int a = 2; a <= n; a++)gg = gcd(gg, tu[a]); if (gg > 1) { cout << "YES" << endl; cout << 0; return 0; } int cz = 0; for (int a = 1; a <= n; a++) { int z = a - 1, y = a + 1; if (tu[a] % 2 == 0)continue; if (y <= n) { if (tu[y] % 2)tu[y] = tu[a] = 2,cz++; else { cz += 2; tu[a] = 2; } continue; } if (z > 0) { cz += 2; tu[a] = 2; } } cout << "YES" << endl; cout << cz;}
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