ACM--最大公约数--HDOJ 1019--Least Common Multiple--水
2016-07-12 14:13
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HDOJ题目地址:传送门
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 45286 Accepted Submission(s): 17030
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
Sample Output
题意题解:求给定你的一排数的最大公约数,这里特别要注意的一点就是需要用__int64,否则会WA
Least Common Multiple
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 45286 Accepted Submission(s): 17030
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
题意题解:求给定你的一排数的最大公约数,这里特别要注意的一点就是需要用__int64,否则会WA
#include<stdio.h> #include<iostream> using namespace std; __int64 lcm(__int64 a,__int64 b){ __int64 temp=a*b; __int64 r=0; while(b){ r=a%b; a=b; b=r; } return temp/a; } int main(){ int i ,n; cin>>n; while(n--){ int t; __int64 a; int b; cin>>t; cin>>a; for(int i=1;i<t;i++){ cin>>b; a=lcm(a,b); } printf("%I64d\n",a); } }
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