zoukankan      html  css  js  c++  java
  • 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


    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:

    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
    ",sum);
                    }else{
                        puts("-1");
                    }
                }
            }
        }
        return 0;
    }
    

      

     
  • 相关阅读:
    几个shell自动化脚本(定期清理、磁盘空间、搜寻关键字)
    linux系统垃圾清理
    mysql出现Got error 28 from storage engine错误
    WebBindingInitializer学习
    Java多线程异步调度程序分析(二)
    自己封装的C#操作redis公共类
    Java多线程编程的常见陷阱(转)
    Java分布式优秀资源集合
    JVM GC之对象生死
    Java内存模型
  • 原文地址:https://www.cnblogs.com/chengsheng/p/4892680.html
Copyright © 2011-2022 走看看