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  • A1059. Prime Factors (25)

    Given any positive integer N, you are supposed to find all of its prime factors, and write them in the format N = p1^k1 * p2^k2 *…*pm^km.

    Input Specification:

    Each input file contains one test case which gives a positive integer N in the range of long int.

    Output Specification:

    Factor N in the format N = p1^k1 * p2^k2 *…*pm^km, where pi's are prime factors of N in increasing order, and the exponent ki is the number of pi -- hence when there is only one pi, ki is 1 and must NOT be printed out.

    Sample Input:

    97532468
    

    Sample Output:

    97532468=2^2*11*17*101*1291

     1 #include <stdio.h>
     2 #include <stdlib.h>
     3 #include <iostream>
     4 #include <string.h>
     5 #include <string>
     6 #include <math.h>
     7 #include <algorithm>
     8 using namespace std;
     9 
    10 
    11 const int maxn=100000;
    12 int prime[maxn],pnum=0;
    13 bool p[maxn]={0};
    14 void Find_Prime(){
    15     for(int i=2;i<maxn;i++)
    16     {
    17         if(p[i]==false)
    18         {
    19             prime[pnum++]=i;
    20             for(int j=i+i;j<maxn;j+=i)
    21             {
    22                 p[j]=true;
    23             }
    24         }
    25     }
    26 } 
    27 
    28 struct factor{
    29     int x,cnt;
    30 }fac[10];
    31 int main(){
    32    Find_Prime();
    33    int n,num=0;
    34    scanf("%d",&n);
    35    if(n==1){ 
    36    printf("1=1");
    37    return 0;
    38    }else{
    39        printf("%d=",n);
    40    }
    41    
    42    
    43    int sqr=(int)sqrt(1.0*n);
    44    int count=0;
    45    for(int i=2;i<=sqr;)
    46    {
    47        if(n%i==0)
    48        {
    49            fac[num].x=i;
    50            fac[num].cnt=0;
    51            while(n%i==0)
    52            {
    53                fac[num].cnt++;
    54                n/=i;
    55            }
    56            num++;    
    57        }
    58        
    59        count++;
    60        i=prime[count];
    61    }
    62    if(n>sqr)
    63    {
    64        fac[num].x=n;
    65        fac[num++].cnt=1;
    66    } 
    67    for(int i=0;i<num;i++)
    68    {
    69        if(i>0)printf("*");
    70        printf("%d",fac[i].x);
    71        if(fac[i].cnt>1)
    72        {
    73            printf("^%d",fac[i].cnt);
    74        }
    75    } 
    76     return 0;
    77 }
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  • 原文地址:https://www.cnblogs.com/ligen/p/4303563.html
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