HDUOJ 1005 Number Sequence(DP求公式)

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 31 2 100 0 0

Sample Output

25

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#includeusing 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#includeint 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]); }}

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