杭电ACM 1019Problem DescriptionThe 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.InputInput will consist of multi

来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/23 21:44:18
杭电ACM1019ProblemDescriptionTheleastcommonmultiple(LCM)ofasetofpositiveintegersisthesmallestpositivei

杭电ACM 1019Problem DescriptionThe 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.InputInput will consist of multi
杭电ACM 1019
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
#include
#include
#include
int gcd(int a,int b)
{
\x05int t=0;
\x05if(a

杭电ACM 1019Problem DescriptionThe 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.InputInput will consist of multi
首先max=b[0]应该加一个条件,就是说要在k=0的时候才行,否则的话,每次都要对他赋值,这个是没有必要的.(if(k==0) max=b[0];)
其次,你的算法有问题,最小公倍数不是两个数相乘就行了,而是能被两者除尽的最小的那个数.
代码如下:
我修改了你和回答人的代码,并且已经A了

#include "stdio.h"
long int LCM(int a,int b)//求最小公倍数的函数
{
long int x;
if (a>b)
{
x=a;
while (x%b!=0)
{ //小优化,既然x是a和b的最小公倍数,所以枚举时只要一直加a或b,再判断能否整除另一个数
x+=a;
}
return x;

}
else
{
x=b;
while (x%a!=0)
{
x+=b;
}
return x;
}
}
int main()
{
int m,n,i,j,k;
long int max,a[100],b[100];
scanf("%d\n",&m);
for(i=0;i<m;i++)
{
scanf("%d",&n);
for(j=0;j<n;j++)
scanf("%d",&a[j]);
for(k=1;k<n;k++)
{
a[k]=LCM(a[k],a[k-1]);//计算相连的两个数的最小公倍数,并替换

}
printf("%ld\n",a[k-1]);

}

return 0;
}

杭电ACM problem 1002 A + B Problem II为什么会WRONG ANSWER?计算结果没错啊?#include 杭电acm 什么思路啊 杭电acm 2035 题的算法是怎样的,杭电acm 2035 题的算法是怎样的,我要算法分析,不要代码!Problem Description求A^B的最后三位数表示的整数.说明:A^B的含义是“A的B次方”Input输入数据包含多个测试实 杭电acm第3809题的详细思路 杭电ACM 3809的详细解题思路是什么 杭电ACM第2136题Largest prime factor, 杭电acm怎么查看自己ac过的代码 杭电 acm 1019!WR!Problem DescriptionThe 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.InputInput will consist of 杭电ACM 1019Problem DescriptionThe 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.InputInput will consist of multi 杭电ACM 2012 不能AC,哪里不对?Problem Description对于表达式n^2+n+41,当n在(x,y)范围内取整数值时(包括x,y)(-39 杭电ACM题1114.把并代码写一下给我,最好再写一下解决的方法说明.我懒得写.貌似挺简单的,帮个忙,谢了Problem DescriptionBefore ACM can do anything, a budget must be prepared and the necessary financial support obtained 杭电acmd 字打错了,是 杭电acm的超时是什么意思 杭电acm 1002 用c或者c++Problem DescriptionI have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B. InputThe first line of the input contains an integer T(1 杭电ACM 1051 求思路好像要用贪心法,能举例更好 杭电acm 1008 题我的为什么是wrong answer 杭电acm题1407,求得最小解,是在x 杭电 acm 3079,怎么做求代码附加讲解,新手, 杭电的acm的题A+B for Input-Output Practice (IV)Problem DescriptionYour task is to Calculate the sum of some integers.InputInput contains multiple test cases.Each test case contains a integer N,and then N integers follow in the same line.A test c