UVA 583 分解质因数

Webster defines prime as:
prime (prim) n. [ME, fr. MF, fem. of prin first, L primus; akin to L prior] 1: first in
time: original 2 a: having no factor except itself and one ⟨3 is a  number⟩ b : having
no common factor except one ⟨12 and 25 are relatively ⟩ 3 a: first in rank, authority or
significance: principal b: having the highest quality or value ⟨ television time ⟩ [from
Webster’s New Collegiate Dictionary]
The most relevant definition for this problem is 2a: An integer g > 1 is said to be prime if and only
if its only positive divisors are itself and one (otherwise it is said to be composite). For example, the
number 21 is composite; the number 23 is prime. Note that the decompositon of a positive number g
into its prime factors, i.e.,
g = f1  f2 


 fn
is unique if we assert that fi > 1 for all i and fi  fj for i < j.
One interesting class of prime numbers are the so-called Mersenne primes which are of the form
2p 􀀀 1. Euler proved that 231 􀀀 1 is prime in 1772 — all without the aid of a computer.
Input
The input will consist of a sequence of numbers. Each line of input will contain one number g in the
range 􀀀231 < g < 231, but different of -1 and 1. The end of input will be indicated by an input line
having a value of zero.
Output
For each line of input, your program should print a line of output consisting of the input number and
its prime factors. For an input number g > 0, g = f1  f2 


 fn, where each fi is a prime number
greater than unity (with fi  fj for i < j), the format of the output line should be
g = f1 x f2 x : : : x fn
When g < 0, if j g j= f1  f2 


 fn, the format of the output line should be
g = -1 x f1 x f2 x : : : x fn
Sample Input
-190
-191
-192
-193
-194
195
196
197
198
199
200
0
Sample Output
-190 = -1 x 2 x 5 x 19
-191 = -1 x 191
-192 = -1 x 2 x 2 x 2 x 2 x 2 x 2 x 3
-193 = -1 x 193
-194 = -1 x 2 x 97
195 = 3 x 5 x 13
196 = 2 x 2 x 7 x 7
197 = 197
198 = 2 x 3 x 3 x 11
199 = 199
200 = 2 x 2 x 2 x 5 x 5

 

直接上代码。

AC code:

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
bool u[111115];
int su[111115];
int num;
void olas()
{
    num=1;
    memset(u,true,sizeof(u));
    for(int i=2; i<=111111; i++)
    {
        if(u[i])    su[num++]=i;
        for(int j=1; j<num; j++)
        {
            if(i*su[j]>111111)    break;
            u[i*su[j]]=false;
            if(i%su[j]==0)    break;
        }
    }
}
int main()
{
    //freopen("input.txt","r",stdin);
    olas();
    ll n;
    while(~scanf("%lld",&n)&&n)
    {
        bool f=true;
        printf("%lld = ",n);
        if(n<0)
        {
            printf("-1");
            n=-n;
            f=false;
        }
        for(int i=1; n>1&&i<num; i++)
        {
            while(n%su[i]==0)
            {
                if(f)
                {
                    printf("%d",su[i]);
                    f=false;
                }
                else
                {
                    printf(" x %d",su[i]);
                }
                n/=su[i];
            }
        }
        if(n!=1)
        {
            if(f)
            {
                printf("%lld",n);
                f=false;
            }
            else
            {
                printf(" x %lld",n);
            }
        }
        printf("\n");
    }
    return 0;
}