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HDU 5505——GT and numbers——————【素数】

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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


Problem Description
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.
 

 

Input
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.

T10001N1000000,1M263.

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.
 

 

Output
For each test case,output an answer.
 

 

Sample Input
3
1 1
1 2
2 4
 

 

Sample Output
0
-1
1
 

 

Source
 

 

题目大意:技术分享

 

解题思路:技术分享

 

#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

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