题目描述
Given is a function
f: 0..N –> 0..N
for a non-negative
N
and a non-negative integer
n
≤
N
. One can construct an infinite sequence
F = f 1(n), f 2(n), … f k(n) …
, where
f k(n)
is defined recursively as follows:
f 1(n) = f(n)
and
f k+1(n)
=
f(f k(n))
.
It is easy to see that each such sequence F is eventually periodic, that is periodic from some point onwards, e.g 1, 2, 7, 5, 4, 6, 5, 4, 6, 5, 4, 6 … . Given non-negative integer N ≤ 11000000 , n ≤ N and f, you are to compute the period of sequence F.
Each line of input contains N, n and the a description of f in postfix notation, also known as Reverse Polish Notation (RPN). The operands are either unsigned integer constants or N or the variable x. Only binary operands are allowed: + (addition), * (multiplication) and % (modulo, i.e. remainder of integer division). Operands and operators are separated by whitespace. The operand % occurs exactly once in a function and it is the last (rightmost, or topmost if you wish) operator and its second operand is always N whose value is read from input. The following function:
2 x * 7 + N %
is the RPN rendition of the more familiar infix
(2*x+7)%N
. All input lines are shorter than 100 characters. The last line of input has
N
equal 0 and should not be processed.
For each line of input, output one line with one integer number, the period of F corresponding to the data given in the input line.
输入
暂无
输出
暂无
样例输入
10 1 x N %
11 1 x x 1 + * N %
1728 1 x x 1 + * x 2 + * N %
1728 1 x x 1 + x 2 + * * N %
100003 1 x x 123 + * x 12345 + * N %
0 0 0 N %
样例输出
1
3
6
6
369
参考代码
#include <stdio.h>
#include <string.h>
#include <assert.h>
char op[101][101];
int i,j,k,n,N,no;
int eval(int x)
{
int i,j;
int stack[101], n=0;
for (i=0;i<no;i++)
{
if (!strcmp(op[i],"x"))
{
stack[n++] = x;
} else if (!strcmp(op[i],"+"))
{
stack[n-2] = (stack[n-2]+stack[n-1])%N;
n--;
} else if (!strcmp(op[i],"*"))
{
stack[n-2] = ((long long)stack[n-2]*stack[n-1])%N;
n--;
} else
{
stack[n++] = atoi(op[i]);
}
}
assert (n == 1);
return stack[0];
}
main()
{
int x,xx;
while (2 == scanf("%d%d",&N,&n)&&N)
{
for (i=0;1 == scanf("%s",op[i]) && strcmp(op[i],"%");i++);
no = i-1;
x = xx = n;
for (j=1;;j++)
{
x = eval(x);
xx = eval(xx);
xx = eval(xx);
if (x == xx)
{
for (k=1;x != (xx=eval(xx));k++);
printf("%dn",k);
break;
}
}
}
}
解析
暂无