HDUOJ 1005 Number Sequence(DP求公式)

                                     Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 146218    Accepted Submission(s): 35530

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
   
   
   
   

 

Author

CHEN, Shunbao

 

Source

ZJCPC2004

 

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我试了一下用递归来解决这问题，看看会不会TLE。。。。结果真的TLE。。。

TLE代码：

#include<iostream>
#include<cstdio>
using namespace std;
int ans;int A,B;
int f(int n){
	if(n==1||n==2)return 1;
	if(n>=3)
	return (A*f(n-1)+B*f(n-2))%7;
}
int main()
{
	int n;
	while(~scanf("%d%d%d",&A,&B,&n))
	{
		if(A==0&&B==0&&n==0)break;
		ans=f(n);
		printf("%d\n",ans);
	}
	return 0;
}

改为DP后AC，注意结果每48个为一个周期。（这个坑爹。。。）

AC代码：

#include<iostream>
#include<cstdio>
int main()
{
	int a,b,n;
	while(~scanf("%d%d%d",&a,&b,&n)){
		if(a==0&&b==0&&n==0)break;
		int arr[48];
        arr[1]=1;
        arr[2]=1;
        for(int i=3;i<48;i++)
            arr[i]=((a*arr[i-1]+b*arr[i-2])%7);
        printf("%d\n",arr[n%48]);

	}
}