渣渣ACM日记——1005-Number Sequence（HDOJ）

   Number Sequence
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 124487    Accepted Submission(s): 30244



Problem Description
A number sequence is defined as follows:


f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.


Given A, B, and n, you are to calculate the value of f(n).


 


Input
The input consists of multiple test cases. Each test case contains 3 integers A, B and n on a single line (1 <= A, B <= 1000, 1 <= n <= 100,000,000). Three zeros signal the end of input and this test case is not to be processed.


 

Output
For each test case, print the value of f(n) on a single line.


 


Sample Input
1 1 3
1 2 10
0 0 0
 


Sample Output
2

5

#include "iostream"
using namespace std;
const int MOD=7;
int f[1000];
int main(){
	int a,b,n,i;
	int result;
	f[1]=1;
	f[2]=1;
	while(cin>>a>>b>>n){
		if(a==0&&b==0&&n==0)
		break;
		if(n<=2){
			cout<<1<<endl;
			continue;
		}	
		for(i=3;i<=n;i++){
			f[i]=(a*f[i-1]+b*f[i-2])%MOD;
		//	cout<<i<<"----"<<f[i]<<endl;
			if(i>4&&f[i]==f[4]&&f[i-1]==f[3])//这里坑爹的要i>4 找了好久错误 
			break;
		}
		if(i<n){
			int tmp=(n-2)%(i-2-2);
			if(tmp!=0)
			result=f[tmp+2];
			else 
			result=f[i-2];
		}
		else
		result=f[n];
		cout<<result<<endl;
	}
	return 0;
}



//超时了 很明显！ 
/*
#include "iostream"
using namespace std;
const int MOD=7;
int main(){
	int a,b,n,i;
	while(cin>>a>>b>>n){
		if(a==0&&b==0&&n==0)
		break;
		if(n<=2){
			cout<<1<<endl;
			continue;
		}
		int last1=1,last2=1,now;
		for(i=3;i<=n;i++){
			now=(a*last2+b*last1)%MOD;
			last1=last2;
			last2=now;
		//	cout<<i<<" ---- "<<now<<endl; 
		//通过观察可以看出 每一行都有属于自己规律 
		}
		cout<<now<<endl;
	}
	return 0;
}
*/

//内存超出  不要用数组 
/*#include "iostream"
using namespace std;
const int MOD=7;
int f[100000005];
int main(){
	int a,b,n,i;
	f[1]=1;
	f[2]=1;
	while(cin>>a>>b>>n){
		if(a==0&&b==0&&n==0)
		break;
		for(i=3;i<=n;i++){
			f[i]=(a*f[i-1]+b*f[i-2])%MOD;
		}
		cout<<f[n]<<endl;
	}
	return 0;
}*/