Lingyuan (seals. CPP/C/PAS)

The value of P (m) is the number of positive factors of m (including 1 and M ).

Returns the minimum value of X that satisfies p (x) = n.

For any positive integer N, there are n = p1 ^ A1 * P2 ^ A2 * P3 ^ A3 *...... * PN ^ An; (PI is a prime number)
Number of factors of N (A1 + 1) * (A2 + 1) * (A3 + 1 )*...... * (An + 1 );

For example, 8 = 2*2*2;
2 ^ 1*3 ^ 1*5 ^ 1 = 30;
Because 8 can also be decomposed into this form: 8 = 2*4; 2 ^ 3*3 ^ 1 = 24;

The correct answer is 24.

Assume there is a number of N = b1 * B2 * B3 *...... * Bi *...... * Bn;
If we break it down into n = b1 * B2 * B3 *...... * (Bj * bi )*...... * B [I-1] * B [I + 1] *... * Bn. The calculated result is smaller than the above formula,
This condition must be met:
P [n-I + 1] ^ Bi * P [n-J + 1] ^ (bj)> P [n-J + 1] ^ (Bj * bi)
Get P [n-J + 1] ^ (bj:
P [n-I + 1] ^ Bi> P [n-J + 1] ^ (Bj * (bi-1 ));
If this condition is met, you can assign BJ to BJ * Bi and assign Bi to 0, which indicates clearing;
Then you can.

 

The Code is as follows:

#include<iostream>#include<cmath>#include<cstdio>using namespace std;int prime[]={0,2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,89,97,101,103,107,109,113,127,131,137,139};int n,i,temp,a[1000]={0},j,ans[10000][2],c[10000];double pro(int x,int y){    double ans=x,t=x;    int c[100],i;c[0]=0;    while (y!=0){c[++c[0]]=y%2;y/=2;}    for (i=c[0]-1;i>=1;i--){        ans*=ans;        if(c[i]==1)ans*=t;    }    return ans;}void clean(){    int i,j;    for(i=1;i<=a[0];i++)        for(j=a[0];j>i;j--)        if(a[i]*a[j]!=0)            if (pro(prime[a[0]-i+1],a[i]-1)>pro(prime[a[0]-j+1],a[j]*(a[i]-1)))            {               a[j]*=a[i]; a[i]=0; break;            }}int main(){    freopen("seals.in","r",stdin);    freopen("seals.out","w",stdout);    scanf("%d",&n);    for    (i=2;i<=sqrt(n);i++)     while (n%i==0) { a[++a[0]]=i; n/=i;}     if (n>1)a[++a[0]]=n;      clean();      int tot=0,t;       for (i=1;i<=a[0];i++)    if (a[a[0]-i+1]>0){         temp=a[a[0]-i+1]-1;           ans[++tot][1]=prime[i];           ans[tot][2]=temp;    }    for (i=1;i<tot;i++) printf("%d^%d*",ans[i][1],ans[i][2]);     printf("%d^%d",ans[tot][1],ans[tot][2]); }

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