Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 683 Accepted Submission(s): 190
题目大意:
解题思路:
#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<iostream>
#include<limits.h>
using namespace std;
typedef unsigned long long UINT;
const int maxn = 1e6+2000;
UINT prime[maxn];
void getprime(){
prime[0] = 1; prime[1] = 1;
for(UINT i = 2; i*i <= maxn-10; i++){
if(!prime[i])
for(int j = i*i; j <= maxn-10 ;j += i){
prime[j] = 1;
}
}
}
int main(){
int T;
getprime();
scanf("%d",&T);
UINT a,b;
while(T--){
scanf("%llu%llu",&a,&b);
if(b < a){
puts("-1");
}else if(b == a){
puts("0");
}else{
if(a == 1 || b%a != 0){
puts("-1");
}else{
b /= a;
int sum = 0;
for(UINT i = 2; i <= maxn-100; i++){
if(prime[i]) continue;
UINT tmp = 1 , times = 0;
if(a % i != 0) continue;
while(a % i == 0){
a /= i;
tmp *= i;
}
while(b % tmp == 0){
b /= tmp;
tmp *= tmp;
times ++;
}
if(b % i == 0){
sum = times+1 > sum? times+1:sum;
while( b % i == 0){
b /= i;
}
}else{
sum = times > sum? times:sum;
}
if(b == 1){
break;
}
}
if(b == 1){
printf("%d\n",sum);
}else{
puts("-1");
}
}
}
}
return 0;
}
HDU 5505——GT and numbers——————【素数】
原文:http://www.cnblogs.com/chengsheng/p/4892680.html