HDU 5505——GT and numbers——————【素数】

GT and numbers

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 683 Accepted Submission(s): 190


[align=left]Problem Description[/align]
You are given two numbers N and M.

Every step you can get a new N in the way that multiply N by a factor of N.

Work out how many steps can N be equal to M at least.

If N can't be to M forever,print −1.

[align=left]Input[/align]
In the first line there is a number T.T is the test number.

In the next T lines there are two numbers N and M.

T≤1000, 1≤N≤1000000,1≤M≤263.

Be careful to the range of M.

You'd better print the enter in the last line when you hack others.

You'd better not print space in the last of each line when you hack others.

[align=left]Output[/align]
For each test case,output an answer.

[align=left]Sample Input[/align]

3

1 1

1 2

2 4

[align=left]Sample Output[/align]

0

-1
1

[align=left]Source[/align]
BestCoder Round #60

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;
}