PAT（顶级）2020年秋季考试 7-2 Number Game (35分)

7-2 Number Game (35分)

 

A number game is to start from a given number A, and to reach the destination number B by a sequence of operations.

For the current number X, there are 3 types of operations:

• X=X+1
• X=X−1
• X=X×N

Your job is to find the minimum number of steps required to reach B from A.

Input Specification:

Each input file contains several test cases. The first line gives a positive integer K (≤10) which is the total number of cases. For each case, three integers are given: A, B and N, where − and 1. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in a line the minimum number of steps required to reach B from A.

Sample Input:

3
3 11 2
-5 -12 3
-2 1000 7

 

Sample Output:

3
2
13

AC代码：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=200005;
bool vis[maxn];
queue<int> que,q;
int handle(int a,int b,int n)
{
    if(a==b) return 0;
    memset(vis,0,sizeof(vis));
    vis[a]=true;
    while(!que.empty()) que.pop();
    while(!q.empty()) q.pop();
    que.push(a);
    q.push(0);
    while(!que.empty())
    {
        int x=que.front(),y=q.front();
        que.pop();q.pop();
        int z=x+1;
        if(z<200000 && vis[z]==false){
            if(z==b) return y+1;
            vis[z]=true;
            que.push(z);
            q.push(y+1);
        }
        z=x-1;
        if(z>0 && vis[z]==false){
            if(z==b) return y+1;
            vis[z]=true;
            que.push(z);
            q.push(y+1);
        }
        z=x*n;
        if(z<200000 && vis[z]==false){
            if(z==b) return y+1;
            vis[z]=true;
            que.push(z);
            q.push(y+1);
        }
    }
}
int main()
{
    int t;
    for(scanf("%d",&t);t;--t){
        int a,b,n,ans=0;
        scanf("%d %d %d",&a,&b,&n);
        if(a==b) ans=0;
        else if(a<b){
            if(a>0) ans=handle(a,b,n);
            else if(a<=0 && b>0) ans=handle(1,b,n)+1-a;
            else ans=b-a;
        }
        else{
            if(a<0) ans=handle(-a,-b,n);
            else if(a>=0 && b<0) ans=handle(1,-b,n)+a+1;
            else ans=a-b;
        }
        printf("%d
",ans);
    }
    return 0;
}