1059. Prime Factors (25)
时间限制
100 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
HE, Qinming
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
找出一个数的所有质数，输入为1等于1，从2开始找,value先排2，再排3，依次往后排，因为是从小到大开始排，所以除的必定是质数，利用cnt判断有多少个，利用flag判断是否为因子，flag1判断第一个用于输出*
代码如下
#include<iostream>
using namespace std;
int main(){
long value;
scanf("%ld",&value);
printf("%ld=",value);
if(value==1) printf("1");
int flag1=0;
for(long i=2;value>=2;++i){
int flag=0;long cnt=0;
while(value%i==0){
++cnt;
flag=1;
value/=i;
}
if(flag==1){
if(flag1==1){
printf("*");
}
printf("%ld",i);
flag1=1;
}
if(cnt>=2){
printf("^%ld",cnt);
}
}
return 0;
}
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