hdu5505(好题)

GT and numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 174    Accepted Submission(s): 53

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.

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.

 

Output

For each test case,output an answer.

 

Sample Input

31 11 22 4

 

Sample Output

0-11

 

Source

BestCoder Round #60

如果$A$大于$B$那么显然无解。

考虑把$A$和$B$分解质因数。

若$B$存在$A$没有的质因数也显然无解。

对于某一个$A$的质因数的次数。为了加速接近$B$，它一定是每次翻倍，最后一次的时候把剩下的加上。

那么答案就是最小的$k$使得2^{k}*A_{num} \geq B_{num}2​k​​∗A​num​​≥B​num​​。

最后把每个质因数的答案max起来即可。

$B$可以刚好等于2^{63}2​63​​，这样就要开unsigned long long。

#include <map>#include <set>#include <stack>#include <queue>#include <cmath>#include <ctime>#include <vector>#include <cstdio>#include <cctype>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>using namespace std;typedef unsigned long long ll;#define INF 0x3f3f3f3f#define inf -0x3f3f3f3f#define lson k<<1, L, mid#define rson k<<1|1, mid+1, R#define mem0(a) memset(a,0,sizeof(a))#define mem1(a) memset(a,-1,sizeof(a))#define mem(a, b) memset(a, b, sizeof(a))int cnt;int prime[1110000];ll is_prime[111000];void isa_prime(){                            //删选素数    mem0(prime);    mem0(is_prime);    cnt=0;    for(int i=2;i<=1000100;i++){        if(prime[i]==0){            is_prime[cnt]=i;            cnt++;            for(int j=i+i;j<=1000100;j+=i)                prime[j]=1;        }    }}int getp(ll x,ll y){    int ans=0;    while(x%y==0&&x>0){        ans++;        x/=y;    }    return ans;}void solve(ll n,ll m){    int ans=1;    int flag=0;    for(int i=0;i<cnt&&is_prime[i]<=n;i++){        if(n%is_prime[i]==0||m%is_prime[i]==0){            int x=getp(n,is_prime[i]);            int y=getp(m,is_prime[i]);            if(x==0&&y!=0){                printf("-1\n");                flag=1;                break;            }            int cnt1=0;            while(y>x){                x*=2;                cnt1++;            }            ans=max(cnt1,ans);            while(n%is_prime[i]==0&&n>0)                n/=is_prime[i];            while(m%is_prime[i]==0&&m>0)//这里要注意                m/=is_prime[i];        }    }    if(flag==0&&m==1)        printf("%d\n",ans);    else if(flag==0)        printf("-1\n");}int main(){    ll n,m;    int t;    isa_prime();    scanf("%d",&t);    for(int case1=1;case1<=t;case1++){        scanf("%I64u%I64u",&n,&m);        if(m%n!=0){            printf("-1\n");            continue;        }        else if(n==1){            if(m==1)                printf("0\n");            else                printf("-1\n");        }        else if(n==m)            printf("0\n");        else{            solve(n,m);        }    }    return 0;}