hdu1005Number Sequence
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Number Sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 108178 Accepted Submission(s): 26289
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).
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#include<stdio.h>#include<math.h>int fun(int a,int b,int n){if(n==1)return 1;else if(n==2)return 1;else return (a*fun(a,b,n-1)+b*fun(a,b,n-2))%7;}int main(){int A,B,N;while(scanf("%d%d%d",&A,&B,&N)!=EOF){if(A==0&&B==0&&N==0)break;else printf("%d\n",fun(A,B,N%49));}return 0;}/*1、根据周期,可以快速获得结果,且周期必定小于49. 周期的原因:如果<f(x),f(x+1)>的值与<f(x+T),f(x+1+T)>一致,那么以后的序列就完全一致了。 周期小于49的原因: 由于mod 7,所以<f(x),f(x+1)>的可能为49个,那么50个这样的<f(x),f(x+1)>必定有两个相同(鸽笼原理),所以最终可以确定周期小于49.2、需要注意的是:并不是整个序列是有周期的,而是其中去掉头部一部分以后是有周期的!!!举个特殊的例子A=7,B=7,那么{f(x)}为{1,1,0,0,0,0,...},其在去掉头部的两个1后是有周期的,所以应该求出这个头部。类似的例子还有{A=2,B=7},{A=3,B=7}...*/
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