poj2115(线性同余方程，类似于poj1061）

C Looooops

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 14918 Accepted: 3766

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 < 2^k) modulo 2^k.

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 < 2^k) 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 2 163 7 2 167 3 2 163 4 2 160 0 0 0

Sample Output

0232766FOREVER设一共执行了X步，则A+CX (mod m)=B (mod m)既 A+CX-B=mY;(想成一个圈,Y个周期)A+CX-mY=BCX+mY=B-A.....应该很像了吧。。。
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;typedef long long ll;void exgcd(ll a,ll b,ll &d,ll &x,ll &y){    if(b==0)    {d=a;x=1;y=0;}    else    {        exgcd(b,a%b,d,y,x);        y-=x*(a/b);    }}int main(){    ll A,B,C,K;    while(scanf("%lld%lld%lld%lld",&A,&B,&C,&K)&&(A+B+C+K))    {        ll x0,y0,d;        ll m=(ll)1<<K;        //cout<<m<<endl;        ll a=C,b=m,c=B-A;        exgcd(a,b,d,x0,y0);        if(c%d!=0)  printf("FOREVER\n");        else        {            x0=c/d*x0;            if(b<0) b=-b;            x0=(x0%(b/d)+b/d)%(b/d);            printf("%lld\n",x0);        }    }    return 0;}