In a strange planet there are n races. They are completely different as well as their food habits. Each race has a food-eating period. That means the ith race eats after every xi de-sec (de-sec is the unit they use for counting time and it is used for both singular and plural). And at that particular de-sec they pass the whole day eating.
The planet declared the de-sec as ‘Eid‘ in which all the races eat together.
Now given the eating period for every race you have to find the number of de-sec between two consecutive Eids.
Input
Input starts with an integer T (≤ 225), denoting the number of test cases.
Each case of input will contain an integer n (2 ≤ n ≤ 1000) in a single line. The next line will contain n integers separated by spaces. The ith integer of this line will denote the eating period for the ith race. These integers will be between 1 and 10000.
Output
For each case of input you should print a line containing the case number and the number of de-sec between two consecutive Eids. Check the sample input and output for more details. The result can be big. So, use big integer calculations.
Sample Input |
Output for Sample Input |
|
2 3 2 20 10 4 5 6 30 60 |
Case 1: 20 Case 2: 60 |
题意分析：求出所有数的最小公倍数，转化为求出每一个数包含的素因子，多个数某一素因子相同取包含这一素因子最多的那一个，最后累乘起来
#include<stdio.h> #include<string.h> #include<math.h> #include<algorithm> #define LL long long using namespace std; bool book[10005]; int pri[2101], s=0, ss; int sum[10005], ans[2001], b[1005]; void inin() { memset(book, true, sizeof(book)); for(int i=2; i<=10000; i++) { if(book[i]){pri[s++]=i; for(int j=i+i; j<=10000; j+=i) book[j]=false; } } } int multiply() { int p=0; ans[p]=1; for(int i=0; i<ss; i++) { int q=0; for(int j=0; j<=p; j++) { int qq=ans[j]*b[i]+q; q=qq/10000; ans[j]=qq%10000;//每四位存一下，节省空间节省时间 } if(q>0) ans[++p]=q; } return p; } int main() { int T, t=1, a; inin(); scanf("%d", &T); while(T--) { ss=0; int n; scanf("%d", &n); memset(sum, 0, sizeof(sum)); for(int i=0; i<n; i++) { scanf("%d", &a); for(int j=0; j<s; j++) { int q=0; while(a%pri[j]==0) { q++; a/=pri[j]; } sum[pri[j]]=max(q,sum[pri[j]]); } } for(int i=0; i<s; i++) { int m=1; for(int k=0; k<sum[pri[i]]; k++) m*=pri[i]; if(m>1)b[ss++]=m; } int mm=multiply(); printf("Case %d: ", t++); printf("%d", ans[mm]); for(int i=mm-1; i>=0; i--)printf("%04d", ans[i]); printf("\n"); } return 0; }
还可以用java大整数搞一搞，经济实惠方便
import java.util.*; import java.io.BufferedInputStream; import java.math.*; public class Main { public static void main(String[] args) { Scanner in=new Scanner(System.in); int T=in.nextInt(); int t=1; while(T-->0) { BigInteger ans=BigInteger.ONE; int n=in.nextInt(); while(n-->0) { BigInteger a=in.nextBigInteger(); ans=ans.multiply(a).divide(ans.gcd(a)); } System.out.print("Case"+" "+(t++)+":"+" "); System.out.println(ans); System.gc(); } in.close(); } }
