http://acm.hdu.edu.cn/showproblem.php?pid=1005A 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
所以对于给定的A和B,可以先打表,找出数列的循环部分. 鸽巢原理知,状态总数不会超过7*7
注意循环节不一定从f(3)开始...
1
#include <iostream>
2using namespace std;
3
4int a,b,n,x,y,l,h,m[7][7],q[300];
5
int main()
{
6
while(scanf("%d%d%d",&a,&b,&n)!=EOF && (a||b||n)){
7
memset(m, 0, sizeof(m));
8 q[1] = q[2] = 1;
9 x = y = 1; l=3;
10 while(m[x][y]==0){ //该状态还未经历过,则扩展
11 q[l] = (a*x+b*y)%7;
12 m[x][y] = l;
13 y = x;
14 x = q[l++];
15
}
16 //此时,q[1h-1]为前面的非循环部分
17 //q[hl-1]为循环节
18 h = m[x][y]; //循环节的起始位置
19 if(n<h) printf("%d\n",q[n]);
20 else printf("%d\n",q[((n-h)%(l-h))+h]);
21 }
22 return 0;
23
}
posted on 2009-04-25 11:55
wolf5x 阅读(887)
评论(0) 编辑 收藏 引用 所属分类:
acm_icpc