标签：enter ant ini lib sample form length nts size
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?+?1in 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 Abeautiful by performing operations described above, and "NO" (without quotes) otherwise.
If the answer was "YES", output the minimal number of moves needed to make sequenceA beautiful.
Example
2
1 1
YES
1
3
6 2 4
YES
0
2
1 3
YES
1
题意：输入n个数 (2?≤?n?≤?100?000），操作:把a[i],a[i+1] 替换成 a[i]-a[i+1],a[i]+a[i+1]，问最少多少次操作
使所有元素的gcd>1。
题解：假设两个数a,b。操作一次a-b,a+b. 操作两次 -2b，2a。gcd=2；
所以任意两个数两次操作后gcd一定>1；
当两个数是偶数时，需要0次操作
当两个是奇数时，需要1次操作
当一奇一偶时，需要2次操作
先循环一遍两个都是奇数的，然后把这两个数更改为偶数，操作次数+1；
在循环一遍一奇一偶成对的，然后把这两个数更改为偶数，操作次数+2；
代码：
#include<iostream> #include<string> #include<algorithm> #include<cstring> #include<cstdlib> #include<cstdio> long long gcd(long long a,long long b) { if(b==0) return a; else gcd(b,a%b); } using namespace std; int main() { long long a[100005],ans; int n,i,num=0; cin>>n; for(i=1;i<=n;i++) cin>>a[i]; ans=gcd(a[1],a[2]); for(i=3;i<=n;i++) ans=gcd(ans,a[i]); if(ans>1) cout<<"YES"<<endl<<0<<endl; else { for(i=1;i<n;i++) { if(a[i]%2&&a[i+1]%2) { a[i]=0; a[i+1]=0; num++; } } for(i=1;i<n;i++) { if(a[i]%2==0&&a[i+1]%2==1) { a[i]=0; a[i+1]=0; num+=2; } else if(a[i]%2==1&&a[i+1]%2==0) { a[i]=0; a[i+1]=0; num+=2; } else continue; } cout<<"YES"<<endl<<num<<endl; } }
标签：enter ant ini lib sample form length nts size
原文地址：http://www.cnblogs.com/GXXX/p/6814992.html
