UPC备战省赛组队训练赛第十七场
with zyd,mxl
G: Greatest Common Divisor
题目描述There is an array of length n, containing only positive numbers.Now you can add all numbers by 1 many times. Please find out the minimum times you need to perform to obtain an array whose greatest common divisor(gcd) is larger than 1 or state that it is impossible.You should notice that if you want to add one number by 1, you need to add all numbers by 1 at the same time.输入The first line of input file contains an integer T (1≤T≤20), describing the number of test cases.Then there are 2×T lines, with every two lines representing a test case.The first line of each case contains a single integer n (1≤n≤1e5) described above.The second line of that contains n integers ranging in [1,1e9].输出You should output exactly T lines.For each test case, print Case d: (d represents the order of the test case) first. Then output exactly one integer representing the answer. If it is impossible, print -1 instead.
样例输入 31 25 2 5 9 5 75 3 5 7 9 11样例输出 Case 1: 0 Case 2: -1 Case 3: 1
题意：
定义a[]数组存储输入的 n 个数；
求使得 ∀i∈[1,n] GCD(a[i]+x) > 1 的最小的 x；
如果不存在这样的x，输出-1；
思路：
将数组 a 排序，去重；
①去重后，如果只有一个元素，输出 (a[1] == 1 ? 1:0)；
②找到相邻两数差值的GCD记为gcd：
(2.1)如果 gcd == 1 ，输出 -1
(2.2)反之，所有数肯定可以通过增加 x 使得所有数变为 gcd 的倍数，当然也可以变为gcd因子的倍数，
求解出最小的x输出；
AC代码：
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define ll long long 4 const int maxn=1e5+50; 5 6 int n; 7 int a[maxn]; 8 9 int GCD(int _a,int _b) 10 { 11 return _a == 0 ? _b:GCD(_b%_a,_a); 12 } 13 ll Solve() 14 { 15 sort(a+1,a+n+1); 16 int t=unique(a+1,a+n+1)-a; 17 t--; 18 if(t == 1)///情况① 19 return a[1] == 1 ? 1:0; 20 21 ll gcd=a[2]-a[1]; 22 for(int i=2;i <= t;++i) 23 gcd=GCD(gcd,a[i]-a[i-1]); 24 25 if(gcd == 1)///情况(2.1) 26 return -1; 27 ll ans=(a[1]/gcd+(a[1]%gcd == 0 ? 0:1))*gcd-a[1];///情况(2.2) 28 for(ll i=2;i*i <= gcd;++i) 29 { 30 if(gcd%i != 0) 31 continue; 32 ll j=gcd/i; 33 ///找到最小的 curAns 使得所有数 +curAns 都可以变为 i,j 的倍数 34 ll curAns=(a[1]/i+(a[1]%i == 0 ? 0:1))*i-a[1]; 35 curAns=min(curAns,(a[1]/j+(a[1]%j == 0 ? 0:1))*j-a[1]); 36 ans=min(ans,curAns); 37 } 38 return ans; 39 } 40 int main() 41 { 42 int test; 43 scanf("%d",&test); 44 for(int kase=1;kase <= test;++kase) 45 { 46 scanf("%d",&n); 47 for(int i=1;i <= n;++i) 48 scanf("%d",a+i); 49 50 printf("Case %d: %lld\n",kase,Solve()); 51 } 52 return 0; 53 }
小结：
1 比赛的时候，只考虑了使所有的数都变成gcd的最小的因子的倍数的情况； 2 并没有考虑到所有数变成gcd的其他因子的倍数使得答案最小； 3 赛后，吃完午饭美美的睡上了一觉； 4 午睡刚醒，就看到队友zyd给我发的G题ac的截图； 5 一脸懵逼的我问了句为啥？？？？ 6 例如 7 13 34 8 差值为21 9 21的非1的因子有3,7,21 10 所有数都变成3的倍数需要 +2 11 所有数都变成7的倍数需要 +1 12 所有数都变成21的倍数需要 +8 13 答案当然是1啦，所以说，最优解不一定是变成gcd最小因子的倍数
