CF 389A:Fox and Number Game
2014-03-06 23:01
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给定n个数,从这n个数中选取任意两个数做差,并把差赋给其中大的数,重复若干次,直到所有数相等,此时求所有数的和。
这个是辗转相减法的原理,所以用辗转相除法求这n个数的最大公约数,然后乘以n即可。
这个是辗转相减法的原理,所以用辗转相除法求这n个数的最大公约数,然后乘以n即可。
#include <cstdio> #include <iostream> using namespace std; int num[105] = {0} ; int gcd(int x , int y) { return (x % y == 0)? y : gcd(y , x % y) ; } int main() { int n ; cin >> n ; for (int i = 0 ; i < n ; i ++) { cin >> num[i] ; } for (int i = 1 ; i < n ; i ++) { if (num[0] > num[i]) num[0] = gcd(num[0] , num[i]) ; else if (num[0] < num[i]) num[0] = gcd(num[i] , num[0]) ; } cout << num[0] * n << endl ; return 0 ; }
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