【数学】【AOJ-41】Least Common Multiple
Description
The least common multiple (LCM) of a set of positive integers is the smallest positive integer which is
divisible by all the numbers in the set. For example, the LCM of 5, 7 and 15 is 105.
divisible by all the numbers in the set. For example, the LCM of 5, 7 and 15 is 105.
Input
Input will consist of multiple problem instances. The first line of the input file will contain a single
integer indicating the number of problem instances. Each instance will consist of a single line of the
form m n1 n2 n3... nm where m is the number of integers in the set and n1 ...nm are the integers. All
integers will be positive and lie within the range of a 32-bit integer.
integer indicating the number of problem instances. Each instance will consist of a single line of the
form m n1 n2 n3... nm where m is the number of integers in the set and n1 ...nm are the integers. All
integers will be positive and lie within the range of a 32-bit integer.
Output
For each problem instance, output a single line containing the corresponding LCM. All results will lie
in the range of a 32-bit integer.
in the range of a 32-bit integer.
Sample Input
Original | Transformed |
2 3 5 7 15 6 4 10296 936 1287 792 1
Sample Output
105 10296
思路:
每两个两个求最小公倍数
参考代码:
#include <stdio.h> int GCD( int a, int b) //最大公约数 { if (!b) return a; return GCD(b,a%b); } int LCM( int a, int b) //最小公倍数 { return a/GCD(a,b)*b; } int a[10000]; int main() { int t; scanf ( "%d" ,&t); while (t--) { int n; scanf ( "%d" ,&n); int i; for (i=0;i<n;i++) scanf ( "%d" ,&a[i]); int temp=a[0]; for (i=1;i<n;i++) temp=LCM(temp,a[i]); printf ( "%d\n" ,temp); } return 0; } |
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