ACM-简单题之Least Common Multiple——hdu1019
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Least Common Multiple
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 28975 Accepted Submission(s): 10905
Problem 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.
Input
Input will consist of multiple problem instances. The first line of the input 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.
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.
Sample Input
2 3 5 7 15 6 4 10296 936 1287 792 1
Sample Output
105 10296
Source
这道题就是求一堆数的最小公倍数。
要注意,每一行的第一个数表示后面跟着有几个数。
然后,我的方法是将输入的数存到数组,然后进行从大到小排序。
从头開始推断,假设当前的mul(之前全部数的最小公倍数)是当前数的倍数,就不须要计算这个数了,直接跳过。
最后再注意一点,用__int64来解决,scanf是用:%I64d
/**************************************** ***************************************** * Author:Tree * *From : blog.csdn.net/lttree * * Title : Least Commn Multiple * *Source: hdu 1019 * * Hint : * ***************************************** ****************************************/ #include <stdio.h> #include <algorithm> using namespace std; __int64 num[10001]; // 求a和b的最小公倍数 __int64 lcm( __int64 a,__int64 b) { __int64 x=a,y=b,k; while( x%y!=0 ) { k=x%y; x=y; y=k; } return a*b/k; } // 从大到小排序 sort用 bool cmp(__int64 a,__int64 b) { return a>b; } int main() { int t,n,i; __int64 mul; scanf("%d",&t); while( t-- ) { scanf("%d",&n); for( i=0;i<n;++i ) scanf("%I64d",&num[i]); sort(num,num+n,cmp); mul=num[0]; // 假设当前的最小公倍数是该数的倍数就不须要计算了。 for( i=1;i<n;++i ) if( mul%num[i]!=0 ) mul=lcm(mul,num[i]); printf("%I64d\n",mul); } return 0; }