2016-2017 CT S03E08: Codeforces Trainings Season 3 Episode 8 K1 gcd
FairWarning
On our planet, Jamcode IX, threeGreat Events occurred. They happened 26000, 11000 and 6000 slarboseconds ago.In 4000 slarboseconds, the amount of time since all of those events will bemultiples of 5000 slarboseconds, the largest possible amount...and theapocalypse will come.
Luckily for you, you live onJamcode X! The apocalypse came on Jamcode IX less than a year ago. But JamcodeX has a worrying prophecy: After the moment of reckoning, on the rst optimumanniversary of the n Great Events,the apocalypse will come. 64 bits will not save you. You have been warned.
The people of Jamcode X are veryconcerned by this prophecy. All of the Great Events have already happened, andtheir times have been measured to the nearest slarbosecond; but nobody knowswhen their optimum anniversary will occur. After studying the diary of ascientist from Jamcode IX, scientists working on the problem have come up witha theory:
The moment of reckoning is now, themoment you solve this problem. At some time y≥ 0 slarboseconds from now, the number of slarboseconds since each of theGreat Events will be divisible by some maximum number t. If you can nd the smallest value of y that gives this largest possible t, that will give you the optimum anniversary when the apocalypsewill come.
On Jamcode IX, for example, therewere 3 Great Events and they happened 26000, 11000 and 6000 slarbosecondsbefore the moment of reckoning. 4000 slarboseconds later, the amount of timesince each event was a multiple of t =5000 slarboseconds, and the apocalypse came.
Your job is to compute the amountof time until the apocalypse comes.
Input
The rst line of the input gives asingle integer n (2 ≤ n ≤ 3).
The second line of the input gives n space-separated integers ti (1 ≤ ti ≤ 108), the number of slarboseconds sinceGreat Event i occurred. It isguaranteed that ti ≠ tj for some i, j.
Output
Output one line containing theminimum number of slarboseconds until ti+y is a multiple of thelargest possible integer factor t forall i.
Examples
| standard input |
standard output |
| 3 26000000 11000000 6000000 |
4000000 |
| 3 1 10 11 |
0 |
| 2 8000001 9000001 |
999999 |
Notes
Epilogue
Fortunately for the peoples of theJamcode system, the apocalypse turned out to be a mistranslation of the giantparty. Nobody from Jamcode IX bothered to pass this along, because they werehaving so much fun.
题意：给你2个或者3个数，找出一个s，使得他们+s后的gcd最大。
题解：
打了表
发现有这样一个规律
对于n个数+s
先排序 然后gcd(a[2]-a[1],a[3]-a[2],...,a[n]-a[n-1])就是他们的最大公约数
所以先找出来 暴力碾过去就行了
对于hard版本 用大数二分查找就能过
ps：如果想优化可以二分找出大于等于a[1]的第一个数 而且这个数是gcd的倍数 然后输出这个数与a[1]的差值即可
#include
#include
#include
#include
using namespace std;
typedef long long ll;
int main(){
ll i,a[4],n;
scanf("%lld",&n);
for(i=1;i<=n;i++)scanf("%lld",&a[i]);
sort(a+1,a+1+n);
if(n==2){
ll now=a[2]-a[1];
for(i=0;;i++){
if((a[1]+i)%now==0)break;
}
printf("%lld\n",i);
}
else{
ll now=__gcd(a[2]-a[1],a[3]-a[2]);
for(i=0;;i++){
if((a[1]+i)%now==0)break;
}
printf("%lld\n",i);
}
return 0;
}