搜搜考试网杭电 acm 1019!WR!
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杭电 acm 1019!WR!
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
代码：
#include
void main()
{
\x05long int n,m,i,a[1000],max,s,j,x;
\x05while(scanf("%ld",&n)!=EOF)
\x05{
\x05\x05while(n--)
\x05\x05{
\x05\x05\x05scanf("%ld",&m);
\x05\x05\x05for(i=0;i
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
代码：
#include
void main()
{
\x05long int n,m,i,a[1000],max,s,j,x;
\x05while(scanf("%ld",&n)!=EOF)
\x05{
\x05\x05while(n--)
\x05\x05{
\x05\x05\x05scanf("%ld",&m);
\x05\x05\x05for(i=0;i
#include
void main()
{
long int n,m,i,a[1000],max,s,j,x;
while(scanf("%ld",&n)!=EOF)
{
while(n--)
{
scanf("%ld",&m);
for(i=0;i
再问： 我都是用C写的。请问用c++就可以AC是为什么？？？
再答： c++兼容c，有些标准c里面不允许的c++可能会允许，所以你用c语言写了都是可以用c++提交的
void main()
{
long int n,m,i,a[1000],max,s,j,x;
while(scanf("%ld",&n)!=EOF)
{
while(n--)
{
scanf("%ld",&m);
for(i=0;i
再问： 我都是用C写的。请问用c++就可以AC是为什么？？？
再答： c++兼容c，有些标准c里面不允许的c++可能会允许，所以你用c语言写了都是可以用c++提交的