C Looooops（扩展欧几里得求模线性方程）

Link：http://poj.org/problem?id=2115

C Looooops

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 20167 Accepted: 5416

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

Source

CTU Open 2004

AC code：

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <vector>#include <string>#include <queue>#include <stack>#include <algorithm>#define PI acos(-1.0)#define LINF 1000000000000000000LL#define eps 1e-8#define LL long long#define MAXN 100010 #define MOD 1000000007using namespace std;const int INF=0x3f3f3f3f;LL exgcd(LL A,LL &x,LL B,LL &y){    LL x1,y1,x0,y0;    x0=1;y0=0;    x1=0;y1=1;    LL r=(A%B+B)%B;    LL q=(A-r)/B;    x=0;y=1;    while(r)    {        x=x0-q*x1;        y=y0-q*y1;        x0=x1;        y0=y1;        x1=x;y1=y;        A=B;B=r;r=A%B;        q=(A-r)/B;    }    return B;}LL ABS(LL x){return x>=0?x:-x;}//求ax=b(mod n)等价于ax+ny=b，故可以这样转化过来 int main(){    long long A,B,C,k,x,y;     while(scanf("%I64d%I64d%I64d%I64d",&A,&B,&C,&k)!=EOF)      {    if(!A&&!B&&!C&&!k)break;     long long a,b,n,d;     a=C;    b=B-A;         n=(long long)1<<k;d=exgcd(a,x,n,y); //求a,n的最大公约数d=gcd(a,n)和方程d=ax+by的系数x、y   if(b%d)printf("FOREVER\n"); else  { x=(x*(b/d))%n;//方程ax=b(mod n)的最小解  x=(x%(n/d)+n/d)%(n/d); //方程ax=b(mod n)的最整数小解  printf("%I64d\n",x);  } /* 不知为何，下面的输出会RE！！！         /*LL x0,y0,flag=0;        if(a<b){flag=a;a=b;b=flag;flag=1;}//默认令a>=b,flag=1说明有交换，则对应的系数x和y最后输出时也要交换后输出         LL d=exgcd(a,x0,b,y0);//返回a、b的最大公约数         //求ax+by=c         if(c%d!=0)//无解         {        printf("FOREVER\n");}else{x0=x0*(c/d);//x的一个解        y0=y0*(c/d);//y的一个解        LL t=y0/(a/d),minx=LINF,miny=LINF,minn=LINF;        for(int i=t-1;i<=t+1;i++)        {        //x的所有解是x=x0+b/d*t,y的所有解是：y=y-a/d*t。            x=x0+b/d*i;            y=y0-a/d*i;            //printf("%d %d\n",x,y);//打印所有解            /* if(ABS(x)+ABS(y)<minn)            {                minn=ABS(x)+ABS(y);                minx=ABS(x);                miny=ABS(y);            }*/           /* if(x>=0)            {            minx=min(minx,x);}            if(y>=0)            {            miny=min(miny,y);}        }        if(flag)printf("%lld\n",miny);        else printf("%lld\n",minx);       /* if(flag)printf("%lld\n",miny);        else printf("%lld\n",minx);*/        //printf("%lld\n",minx);  //}    }    return 0;}