hdu 1005 用mod的循環性質

http://acm.hdu.edu.cn/showproblem.php?pid=1005
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).
1 <= A, B <= 1000, 1 <= n <= 100,000,000

解:
f(n) = (A * f(n - 1) + B * f(n - 2)) %7
      = (A * f(n - 1) %7 + B * f(n - 2) %7) %7
所以對於給定的A和B,可以先打表,找出數列的循環部分. 鴿巢原理知,狀態總數不會超過7*7
注意循環節不一定從f(3)開始...
#include <iostream> using namespace std; int a,b,n,x,y,l,h,m[7][7],q[300]; int main() { while(scanf("%d%d%d",&a,&b,&n)!=EOF && (a||b||n)) { memset(m, 0, sizeof(m)); q[1] = q[2] = 1; x = y = 1; l=3; while(m[x][y]==0) { //該狀態還未經歷過,則擴展 q[l] = (a*x+b*y)%7; m[x][y] = l; y = x; x = q[l++]; } //此時,q[1...h-1]爲前面的非循環部分 //q[h...l-1]爲循環節 h = m[x][y]; //循環節的起始位置 if(n<h) { printf("%d/n",q[n]); } else { printf("%d/n",q[((n-h)%(l-h))+h]); } } return 0; }