HDU 5505 - BestCoder Round #60 - GT and numbers

题目链接 : http://bestcoder.hdu.edu.cn/contests/contest_chineseproblem.php?cid=641&pid=1002

思路 :

N有若干个质因子, N = a^b * c^d * e^f......

M也有若干个质因子, M = a^(b+k) * c(d+k1) * e^(f+k2)......

N能到达M的条件是它们的质因子必须完全相同

N每次可以乘上它的若干个质因子, 直到这个质因子的幂次等于M这个质因子的幂次

考虑这样一个事实, N乘上某个质因子的a次幂后, 新的N可以乘上该质因子的2a次幂

故对于每个质因子, N的次数每次平方, cnt++, 直到大于等于M这个质因子的次数, 取最大的cnt即可

要注意的是M的范围会爆long long 所以要开unsigned long long

#include <cstdio>
#include <cstring>
#include <algorithm>
 
using namespace std;
 
typedef unsigned long long LL;
 
const int MAXN = 1e6+;
const int PRI_NUM = 2e6+;
 
bool vis[PRI_NUM];
int prime[MAXN];
int factor_n[MAXN];
int factor_m[MAXN];
int prime_n[];
int prime_m[];
 
void Pre()
{
    for(int i = ; i <= PRI_NUM; i++) {
        if(vis[i] == ) {
            for(int j = i+i; j <= PRI_NUM; j+=i) {
                vis[j] = ;
            }
        }
    }
    int cnt = ;
    for(int i = ; i <= PRI_NUM; i++) {
        if(vis[i] == ) {
            prime[cnt++] = i;
        }
    }
}
 
void Init()
{
    memset(prime_n, , sizeof(prime_n));
    memset(prime_m, , sizeof(prime_m));
}
 
int main()
{
    Pre();
 
    int t;
    int n;
    LL m;
 
    scanf("%d", &t);
    while(t--) {
        Init();
        scanf("%d %I64u", &n, &m);
        if(m == n) {
            printf("0\n");
            continue;
        }
        if(m % n || n <= ) {
            printf("-1\n");
            continue;
        }
        int cnt = ;
        for(int i = ; i < MAXN; i++) {
            if(n == ) break;
            if(n % prime[i] == ) {
                factor_n[cnt] = prime[i];
                while(n % prime[i] == ) {
                    n /= prime[i];
                    prime_n[cnt]++;
                }
                cnt++;
            }
        }
        if(n > ) {
            factor_n[cnt] = n;
            prime_n[cnt] = ;
            cnt++;
        }
        for(int i = ; i < cnt; i++) {
            while(m % factor_n[i] == ) {
                m /= factor_n[i];
                prime_m[i]++;
            }
        }
        if(m > ) {
            printf("-1\n");
            continue;
        }
        int ans = ;
        for(int i = ; i < cnt; i++) {
            int k = ;
            while(prime_n[i] < prime_m[i]) {
                prime_n[i] <<= ;
                k++;
            }
            if(k > ans) ans = k;
        }
        printf("%d\n", ans);
    }
 
    return ;
}

另外看到一种很强的做法

#include <cstdio>
#include <cstring>
#include <algorithm>
 
using namespace std;
 
typedef unsigned long long ULL;
 
ULL Gcd(ULL a, ULL b)
{
    ULL r;
    while(a % b) {
        r = a % b;
        a = b;
        b = r;
    }
    return b;
}
 
int main()
{
    int t;
    ULL n, m;
 
    scanf("%d", &t);
    while(t--) {
        scanf("%I64u %I64u", &n, &m);
        int ans = ;
        bool flag = ;
        while(n != m) {
            if(m % n) {
                flag = ;
                break;
            }
            ULL temp = Gcd(m / n, n);
            if(temp == ) {
                flag = ;
                break;
            }
            n *= temp;
            ans++;
        }
        if(flag == ) {
            printf("%d\n", ans);
        }
        else printf("-1\n");
    }
 
    return ;
}