练习cf1674A. Number Transformation

题目如下
A. Number Transformation
time limit per test2 seconds
memory limit per test512 megabytes
You are given two integers 𝑥 and 𝑦. You want to choose two strictly positive (greater than zero) integers 𝑎 and 𝑏, and then apply the following operation to 𝑥 exactly 𝑎 times: replace 𝑥 with 𝑏⋅𝑥.

You want to find two positive integers 𝑎 and 𝑏 such that 𝑥 becomes equal to 𝑦 after this process. If there are multiple possible pairs, you can choose any of them. If there is no such pair, report it.

For example:

if 𝑥=3 and 𝑦=75, you may choose 𝑎=2 and 𝑏=5, so that 𝑥 becomes equal to 3⋅5⋅5=75;
if 𝑥=100 and 𝑦=100, you may choose 𝑎=3 and 𝑏=1, so that 𝑥 becomes equal to 100⋅1⋅1⋅1=100;
if 𝑥=42 and 𝑦=13, there is no answer since you cannot decrease 𝑥 with the given operations.
Input
The first line contains one integer 𝑡 (1≤𝑡≤104) — the number of test cases.

Each test case consists of one line containing two integers 𝑥 and 𝑦 (1≤𝑥,𝑦≤100).

Output
If it is possible to choose a pair of positive integers 𝑎 and 𝑏 so that 𝑥 becomes 𝑦 after the aforementioned process, print these two integers. The integers you print should be not less than 1 and not greater than 109 (it can be shown that if the answer exists, there is a pair of integers 𝑎 and 𝑏 meeting these constraints). If there are multiple such pairs, print any of them.

If it is impossible to choose a pair of integers 𝑎 and 𝑏 so that 𝑥 becomes 𝑦, print the integer 0 twice.

题目大意
现有x，y，能否找出任意a，b，使得x * （b^a）= y；如果有就输出a，b的值，否则输出“0 0”。

题目分析
根据x * （b^a）= y，可以得到 b^a = y / x，所以可以通过暴力遍历得出a和b的值。
首先要判断x，y是否满足条件；

点击查看代码

 if(y < x || y % x != 0){
            printf("0 0\n");
        }else if(y == x){
            printf("1 1\n");
        }else{
            int n = y / x;
            int a = 0, b = 0;
            if(check(n, &a, &b)){
                printf("%d %d\n", a, b);
            }else{
                printf("0 0\n");
            }
        }

再考虑a，b的可能； 注意，考虑到会出现a等于1的情况。

点击查看代码

int check(int n, int* aa, int* bb){
    for(int b = 2; b * b <= n; b++){
        long long power = b;
        int a = 1;
        while(power <= n){
            if(power == n){
                *aa = a;
                *bb = b;
                return 1;
            }
            power *= b;
            a++;
        }
    }
    for(int b = 2; b <= n; b++){
        if(b == n){
            *aa = 1;
            *bb = b;
            return 1;
        }
    }
    return 0;
}

完整代码

点击查看代码

#include <stdio.h>

int check(int n, int* aa, int* bb){
    for(int b = 2; b * b <= n; b++){
        long long power = b;
        int a = 1;
        while(power <= n){
            if(power == n){
                *aa = a;
                *bb = b;
                return 1;
            }
            power *= b;
            a++;
        }
    }
    for(int b = 2; b <= n; b++){
        if(b == n){
            *aa = 1;
            *bb = b;
            return 1;
        }
    }
    return 0;
}

int main(){
    int t;
    scanf("%d", &t);
    while(t--){
        int x, y;
        scanf("%d%d", &x, &y);
        if(y < x || y % x != 0){
            printf("0 0\n");
        }else if(y == x){
            printf("1 1\n");
        }else{
            int n = y / x;
            int a = 0, b = 0;
            if(check(n, &a, &b)){
                printf("%d %d\n", a, b);
            }else{
                printf("0 0\n");
            }
        }
    }
    return 0;
}