POJ 2115 C Looooops 扩展欧几里德-CFANZ编程社区

C Looooops 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. 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 Source CTU Open 2004

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 31879		Accepted: 9269

算法分析：

扩展欧几里德：给你A，B, C,K, 这个循环的次数for (variable = A; variable != B; variable += C)，即（A+xC）%2^k=B，相当于求（A+XC）+2^kY=B,即（、XC+2^kY=B-A。

代码实现：

#include<cstdio>  
#include<cstring>  
#include<cstdlib>  
#include<cctype>  
#include<cmath>  
#include<iostream>  
#include<sstream>  
#include<iterator>  
#include<algorithm>  
#include<string>  
#include<vector>  
#include<set>  
#include<map>  
#include<stack>  
#include<deque>  
#include<queue>  
#include<list>  
using namespace std;
typedef long long ll;
ll e_gcd (ll a, ll b, ll& x, ll& y)
{
    if (b == 0)
    {
        x = 1, y = 0;
        return a;
    }
    ll ans = e_gcd (b, a % b, y, x);
    y -= a / b * x;
    return ans;
}
int main()
{
    ll a,b,c,k,x0,y0;
    while(scanf("%lld%lld%lld%lld",&a,&b,&c,&k)!=EOF)
    {
        
        
        if(a==0&&b==0&&k==0&&c==0) break;
        if(a==b){cout<<0<<endl;continue;}
        ll temp=a,temp2=b;
        a=c;b=pow(2,k);c=temp2-temp;  //竟然败在这里，一开始写出c=b-temp
        
        ll gcd=e_gcd(a,b,x0,y0);
        
       if(c%gcd)  printf("FOREVER\n");
       else 
        {
            ll t=b/gcd;
            x0*=c/gcd;
            printf("%lld\n",(x0%t+t)%t);
        }

        
    }
    return 0;
    
}