C Looooops
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 19536 | Accepted: 5204 |
Description
A Compiler Mystery: We are given a C-language style for loop of type
for (variable = A; variable != B; variable += C) statement;
I.e., a loop which starts by setting variable to value A and while variable is not equal to B, repeats statement followed by increasing the variable by C. We want to know how many times does the statement get executed for particular values of A, B and C, assuming
that all arithmetics is calculated in a k-bit unsigned integer type (with values 0 <= x < 2k) modulo 2k.
Input
The input consists of several instances. Each instance is described by a single line with four integers A, B, C, k separated by a single space. The integer k (1 <= k <= 32) is the number of bits of the control variable of the loop and A, B, C (0 <= A, B, C
< 2k) are the parameters of the loop.
The input is finished by a line containing four zeros.
< 2k) are the parameters of the loop.
The input is finished by a line containing four zeros.
Output
The output consists of several lines corresponding to the instances on the input. The i-th line contains either the number of executions of the statement in the i-th instance (a single integer number) or the word FOREVER if the loop does not terminate.
Sample Input
3 3 2 16
3 7 2 16
7 3 2 16
3 4 2 16
0 0 0 0
Sample Output
0
2
32766
FOREVER
题意是问在
for (variable = A; variable != B; variable += C)
这样的情况下，循环多少次。
当中全部的数要mod 2的k次方。所以方程就是(A+C*x)%(2^k)=B，变换一下就是-C*x+(2^k)*y=A-B。解这个方程的最小正数x就可以。
又是扩展欧几里德。
代码：
#include <iostream>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std; long long yue; void ex_gcd(long long a,long long b, long long &xx,long long &yy)
{
if(b==0)
{
xx=1;
yy=0;
yue=a;
}
else
{
ex_gcd(b,a%b,xx,yy); long long t=xx;
xx=yy;
yy=t-(a/b)*yy;
}
} int main()
{
long long A,B,C,k,k2,xx,yy; while(scanf_s("%lld%lld%lld%lld",&A,&B,&C,&k))
{
if(!A&&!B&&!C&&!k)
break; k2=(1LL<<k);
ex_gcd(-C,k2,xx,yy); if((A-B)%yue)
{
cout<<"FOREVER"<<endl;
}
else
{
xx=xx*((A-B)/yue);
long long r=k2/yue;
if(r<0)
xx=(xx%r-r)%r;
else
xx=(xx%r+r)%r;
printf("%lld\n",xx);
}
}
return 0;
}