[ABC446E] Multiple-Free Sequences 题解

AT_abc446_e [ABC446E] Multiple-Free Sequences

题目描述

Among integer pairs $ (x, y) $ satisfying $ 0 \leq x,y \leq M-1 $ , how many are there such that the infinite sequence $ (s_1, s_2, \dots) $ defined by the following recurrence relation contains no multiples of $ M $ ?

• $ s_1 = x $
• $ s_2 = y $
• $ s_n = A s_{n-1} + B s_{n-2} $ ( $ n \geq 3 $ )

输入格式

The input is given from Standard Input in the following format:

$ M $ $ A $ $ B $

输出格式

Output the answer on one line.

输入输出样例 #1

输入 #1

4 1 2

输出 #1

7

输入输出样例 #2

输入 #2

446 1 1

输出 #2

0

输入输出样例 #3

输入 #3

1000 784 385

输出 #3

995373

说明/提示

Sample Explanation 1

The integer pairs satisfying the condition in the problem statement are $ (x,y) = (1,1), (1,3), (2,1), (2,2), (2,3), (3,1), (3,3) $ , for a total of seven pairs.

For example, when $ (x,y) = (2,1) $ , the corresponding sequence is $ (2,1,5,7,17,31,65,127,\dots) $ . This sequence contains no multiples of $ 4 $ . Thus, $ (x,y) = (2,1) $ satisfies the condition in the problem statement.

On the other hand, when $ (x,y) = (3,2) $ , the corresponding sequence is $ (3,2,8,12,28,52,108,212,\dots) $ . The third term of this sequence is $ 8 $ , which is a multiple of $ 4 $ . Thus, $ (x,y) = (3,2) $ does not satisfy the condition in the problem statement.

Sample Explanation 2

No integer pairs satisfy the condition in the problem statement.

Constraints

• $ 2 \leq M \leq 1000 $
• $ 0 \leq A, B \leq M-1 $
• All input values are integers.

思路

MAP，直接AC。

代码见下

using namespace std;
long long m,a,b,s[100005],d=0,bo=0,op=0,os=100000;
unordered_map<long long,int> mp;
unordered_map<long long,int> mps;
int main(){
	cin>>m>>a>>b;
//	if(a==1&&b==1){
//		cout<<0<<endl;
//		return 0;
//	}
	for(int i=1;i<=m-1;i++){
		for(int j=1;j<=m-1;j++){
			d=2;
			s[1]=i;
			s[2]=j;
			mp.clear();
			if(mps[i*os+j]==2){
				op++;
				continue;
			}
			if(mps[i*os+j]==1){
				continue;
			}
			mp[i*os+j]=1;
			bo=0;
			for(int k=3;;k++){
				s[k]=(a*s[k-1]+b*s[k-2])%m;
				if(s[k]==0){
					bo=1;
					for(int o=2;o<=k;o++){
						mps[s[o-1]*os+s[o]]=1;
					}
					bo=1;	
					break;					
				}
				if(mps[s[k-1]*os+s[k]]==1){
					for(int o=2;o<=k;o++){
						mps[s[o-1]*os+s[o]]=1;
					}
					bo=1;
					break;
				}		
				if(mps[s[k-1]*os+s[k]]==2){
					for(int o=2;o<=k;o++){
						mps[s[o-1]*os+s[o]]=2;
					}
					bo=0;
					break;
				}				
				if(mp[s[k-1]*os+s[k]]==1){
					if(bo==1){
						for(int o=2;o<=k;o++){
							mps[s[o-1]*os+s[o]]=1;
						}
						bo=1;						
					}
					else{
						for(int o=2;o<=k;o++){
							mps[s[o-1]*os+s[o]]=2;
						}
						bo=0;						
					}
					break;
				} 
				mp[s[k-1]*os+s[k]]=1;
			}
			if(bo==0){
				op++;
			}
		}
	}
	cout<<op<<endl;
	return 0; 	
}