Fox Ciel is playing a game with numbers now.
Ciel has n positive integers: x1, x2, ..., xn. She can do the following operation as many times as needed: select two different indexes i and jsuch that xi > xj hold, and then apply assignment xi = xi - xj. The goal is to make the sum of all numbers as small as possible.
Please help Ciel to find this minimal sum.
The first line contains an integer n (2 ≤ n ≤ 100). Then the second line contains n integers: x1, x2, ..., xn (1 ≤ xi ≤ 100).
Output a single integer — the required minimal sum.
2
1 2
2
3
2 4 6
6
2
12 18
12
5
45 12 27 30 18
15
In the first example the optimal way is to do the assignment: x2 = x2 - x1.
In the second example the optimal sequence of operations is: x3 = x3 - x2, x2 = x2 - x1.
题意:给出n个数字,问任意2个数,如果ai>aj,那么ai=ai-aj。问最后最少总和是多少
思路:就是求最大公约数*n
1 #include<bits/stdc++.h> 2 using namespace std; 3 4 int a[102]; 5 int main(){ 6 int n; 7 cin>>n; 8 for(int i=1;i<=n;i++) cin>>a[i]; 9 for(int i=2;i<=n;i++){ 10 a[i]=__gcd(a[i],a[i-1]); 11 } 12 cout<<a[n]*n<<endl; 13 }