标签：his integer ret index using ini possible ble tps
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).
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 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.
2
1 1
YES
1
3
6 2 4
YES
0
2
1 3
YES
1
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.
由题目给出操作可知：当gcd(a,b)<=1时，进行操作为：
初始：a b
第一步：a-b a+b
第二步：-2b 2a
即两个数最多2步操作就能满足GCD==2。
对于两个偶数，要进行0步操作；对于两个奇数，要进行1步操作；对于一个奇数一个偶数，要进行2步操作。
先把所有“2个奇数成对”的情况计数+1并把两个奇数更新为偶数，然后在重新判断所有“1个奇数1个偶数成对”的情况计数+2。
#include<bits/stdc++.h> using namespace std; typedef long long ll; ll a[200050],n,num=0; ll gcd(ll a,ll b){ return b==0?a:gcd(b,a%b); } int main(){ cin>>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) cout<<"YES"<<endl<<0<<endl; else { for(int i=1;i<n;i++) if(a[i]%2&&a[i+1]%2) a[i]=0,a[i+1]=0,num++; for(int i=1;i<n;i++) if((a[i]%2&&a[i+1]%2==0)||(a[i]%2==0&&a[i]%2)) a[i]=0,a[i+1]=0,num+=2; cout<<"YES"<<endl<<num<<endl; } return 0; }
CodeForce-798C Mike and gcd problem（贪心）
标签：his integer ret index using ini possible ble tps
原文地址：http://www.cnblogs.com/YingZhixin/p/6803484.html
