离异征婚交友网,童话故事mp3,宠奴的逆袭
time limit: 2000/1000 ms (java/others) memory limit: 65536/32768 k (java/others)total submission(s): 64855 accepted submission(s): 24737
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
这道水题连续提交五次，第六次才ac。。。。。。（我真是菜，最近做做的题有点自闭）
这是是我多次修改仍然不ac的题解（测试样例全过自己测了几个也没问题，很绝望）
我要评论#include <stdio.h>
#include <math.h>
int main() {
int t,temp;
scanf("%d",&t);
for(int i=1; i <= t; i++) {
int n;
scanf("%d",&n);
for(int j=1; j <= n ; j++) {
int num;
scanf("%d",&num);
if(j == 1) {
temp=num;
continue;
}
int max;
if(temp > num)
max=num;
else
max=temp;
for( ; max >= 1; max--)
if(temp%max == 0&&num%max == 0)
break;
temp=temp*num/max;
}
printf("%d\n",temp);
}
return 0;
}
最后有dalao指点，把求公因数的方法改成辗转相乘法并用函数嵌套终于ac 无奈绝望╮(╯﹏╰）╭ 关于辗转相除法求最大公因数，举个栗子： a=6, b=4; 第一步 a=a%b=6%4=2； 第二步 b=4； 第三步 a%b=2%4=0，此时a == 0，b=(上一步)a; 此时b就是最大公因数； （算了上详细讲解链接 ）
以下为正确代码
#include <stdio.h>
#include <math.h>
int gcd(int a,int b){ //辗转相除法
if(b == 0)
return a;
return gcd(b,a%b);
}
int lcm(int a,int b)
{
return a/gcd(a,b)*b; //把函数打包
}
int main() {
int t,temp,n;
scanf("%d",&t);
for(int i=1; i <= t; i++) {
scanf("%d",&n);
int temp = 1;
for(int j=1; j <= n ; j++) {
int num;
scanf("%d",&num);
temp = lcm(temp,num);
}
printf("%d\n",temp);
}
return 0;
}
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