Lightoj1024-eid (number of minimum conventions for n number + high precision)

Title Link: http://lightoj.com/volume_showproblem.php?problem=1024

Test instructions: Give you the number of n (2<=n<=1000), and then the number of n least common multiple, the size of each number is 1---10000, so the answer will be very large, may reach 1000 4-digit multiplication, so the result is very large, nearly 4,000 bits;

So it's going to involve high-precision operations, and we can't just cycle through the least common multiple; we can divide a number into multiple prime numbers, and then find the one that corresponds to the most prime numbers in each of them, and then multiply the results.

Example of a sample

4

5 6 30 60

5:5//Description There must be a factor in the least common multiple 5

6:2*3//Description The factor of least common multiple must have a 2 and a 3;

30:2*3*5//Description The factor of least common multiple must have a 2 and a 3 and a 5;

60:2^2*3*5//Description The least common multiple factor must have 2 2 and one 3 and one 5;

So we can ignore those numbers less, find out the results of the description must contain 2 2 a 3 a 5;

It is important to note that because each time a large integer is multiplied by n, so that the tle is not several times, so can be used to output a number of methods;

#include <stdio.h>#include<string.h>#include<algorithm>#include<math.h>typedefLong LongLL;#defineN 10001using namespacestd;Const DoubleEPS = 1e-6;intf[101], p[101], k =0;voidPrime () { for(intI=2; i<101; i++)    {        if(!f[i]) p[k++] =i;  for(intJ=i; j<101; j+=i) f[j]=1; }}intans[1001], cnt[n];voidMul (intA[],intnum) {     for(intI=0; i< +; i++) A[i]= a[i]*num;  for(intI=0; i< +; i++) {A[i+1] + = a[i]/10000; A[i]= a[i]%10000; }}voidPUTS (inta[]) {    intI= +;  while(i>=0&& a[i]==0) i--; printf ("%d", a[i--]);  while(i>=0) printf ("%04d", a[i--]); printf ("\ n");}intMain () {Prime (); intN, t, t =1; scanf ("%d", &T);  while(t--) {memset (CNT,0,sizeof(CNT)); memset (ans,0,sizeof(ans)); intnum; scanf ("%d", &N);  for(intI=1; i<=n; i++) {scanf ("%d", &num);  for(intj=0; J<k && p[j]*p[j] <= num; J + +)            {                ints =0;  while(Num%p[j] = =0) {s++; Num/=P[j]; } Cnt[p[j]]=Max (Cnt[p[j]], s); }            if(Num >1) Cnt[num]= Max (Cnt[num],1); } ans[0] =1;  for(intI=0; i<n; i++)        {            if(!cnt[i])Continue; intRET =1;  for(intj=0; j<cnt[i]; J + +) ret*=i;        Mul (ans, ret); } printf ("Case %d:", t++);    PUTS (ANS); }    return 0;}

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Lightoj1024-eid (number of minimum conventions for n number + high precision)