HDOJ 1005 Number Sequence（规律）

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 127173    Accepted Submission(s): 30910

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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不找循环长度的AC：

#include<stdio.h>int main(){int i,n,A,B;int f[100],num;while(scanf("%d%d%d",&A,&B,&n)!=EOF && (A || B || n)){f[1] = 1;f[2] = 1;for(i=3; i<=99; i++){f[i] = (A * f[i-1] + B * f[i-2]) % 7;if(f[i] == f[i-1] && f[i] == 1) //判断是否出现周期break;}num = i-2;n = n % num;if( n == 0)printf("%d\n",f[num]);elseprintf("%d\n",f[n]);}return 0;}

循环长度49：对于公式 f[n] = A * f[n-1] + B * f[n-2]; 后者只有7 * 7 = 49 种可能，为什么这么说，因为对于f[n-1] 或者 f[n-2] 的取值只有 0,1,2,3,4,5,6 这7个数，A，B又是固定的，所以就只有49种可能值了。

#include<stdio.h>int fun(int a,int b,__int64 n){if(n==1||n==2) return 1;else return(a*fun(a,b,n-1)+b*fun(a,b,n-2))%7;}int main(){int a,b;__int64 n;while(scanf("%d%d%I64d",&a,&b,&n)&&(a||b||n)){printf("%d\n",fun(a,b,n%49));}return 0;}