最小公倍数
时间限制：1000 ms | 内存限制：65535 KB
难度：3
- 描述
-
为什么1小时有60分钟，而不是100分钟呢？这是历史上的习惯导致。但也并非纯粹的偶然：60是个优秀的数字，它的因子比较多。事实上，它是1至6的每个数字的倍数。即1,2,3,4,5,6都是可以除尽60。我们希望寻找到能除尽1至n的的每个数字的最小整数m.
- 输入
- 多组测试数据（少于500组）。
每行只有一个数n(1<=n<=100). - 输出
- 输出相应的m。
- 样例输入
-
2 3 4
- 样例输出
-
2 6 12
//打表
import java.math.BigDecimal;
import java.math.BigInteger;
import java.util.Scanner;
public class Main{
public static void main(String args[]){
Scanner cin = new Scanner(System.in);
/*final int MAX = 105;
int arr[] = new int[MAX];
BigInteger res[] = new BigInteger[MAX];
for(int i=1; i<MAX; ++i)arr[i] = i;
for(int i=2; i<MAX; ++i){
for(int j=i+1; j<MAX; ++j){
if(j%i == 0)
arr[j] /= arr[i];
}
}
for(int i=1; i<MAX; ++i)res[i] = BigInteger.ONE;
for(int i=2; i<MAX; ++i){
for(int j=2; j<i; ++j){
res[i] = res[i].multiply(BigInteger.valueOf(arr[j]));
}
}
for(int i=1; i<101; ++i){
int n = i;
System.out.println("\""+res[n+1] + "\",");
}*/
String s[] = {
"1",
"2",
"6",
"12",
"60",
"60",
"420",
"840",
"2520",
"2520",
"27720",
"27720",
"360360",
"360360",
"360360",
"720720",
"12252240",
"12252240",
"232792560",
"232792560",
"232792560",
"232792560",
"5354228880",
"5354228880",
"26771144400",
"26771144400",
"80313433200",
"80313433200",
"2329089562800",
"2329089562800",
"72201776446800",
"144403552893600",
"144403552893600",
"144403552893600",
"144403552893600",
"144403552893600",
"5342931457063200",
"5342931457063200",
"5342931457063200",
"5342931457063200",
"219060189739591200",
"219060189739591200",
"9419588158802421600",
"9419588158802421600",
"9419588158802421600",
"9419588158802421600",
"442720643463713815200",
"442720643463713815200",
"3099044504245996706400",
"3099044504245996706400",
"3099044504245996706400",
"3099044504245996706400",
"164249358725037825439200",
"164249358725037825439200",
"164249358725037825439200",
"164249358725037825439200",
"164249358725037825439200",
"164249358725037825439200",
"9690712164777231700912800",
"9690712164777231700912800",
"591133442051411133755680800",
"591133442051411133755680800",
"591133442051411133755680800",
"1182266884102822267511361600",
"1182266884102822267511361600",
"1182266884102822267511361600",
"79211881234889091923261227200",
"79211881234889091923261227200",
"79211881234889091923261227200",
"79211881234889091923261227200",
"5624043567677125526551547131200",
"5624043567677125526551547131200",
"410555180440430163438262940577600",
"410555180440430163438262940577600",
"410555180440430163438262940577600",
"410555180440430163438262940577600",
"410555180440430163438262940577600",
"410555180440430163438262940577600",
"32433859254793982911622772305630400",
"32433859254793982911622772305630400",
"97301577764381948734868316916891200",
"97301577764381948734868316916891200",
"8076030954443701744994070304101969600",
"8076030954443701744994070304101969600",
"8076030954443701744994070304101969600",
"8076030954443701744994070304101969600",
"8076030954443701744994070304101969600",
"8076030954443701744994070304101969600",
"718766754945489455304472257065075294400",
"718766754945489455304472257065075294400",
"718766754945489455304472257065075294400",
"718766754945489455304472257065075294400",
"718766754945489455304472257065075294400",
"718766754945489455304472257065075294400",
"718766754945489455304472257065075294400",
"718766754945489455304472257065075294400",
"69720375229712477164533808935312303556800",
"69720375229712477164533808935312303556800",
"69720375229712477164533808935312303556800",
"69720375229712477164533808935312303556800",
};
while(cin.hasNext()){
int n = cin.nextInt();
System.out.println(s[n-1]);
}
}
}