POJ 2115 C Looooops(拓展欧几里得)

C Looooops
Time Limit: 1000MS      Memory Limit: 65536K
Total Submissions: 24296        Accepted: 6765
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

这道题题意很容易理解:
其实它就是求A+Cx≡B(mod 2^k )
这里就用拓展欧几里得求x的最小非负整数啦。
记得数据类型就行啦。

#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cctype>
#include<cmath>
#include<ctime>
#include<string>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#include<set>
#include<map>
#include<cstdio>
#include<limits.h>
#define MOD 1000000007
#define fir first
#define sec second
#define fin freopen("/home/ostreambaba/文档/input.txt", "r", stdin)
#define fout freopen("/home/ostreambaba/文档/output.txt", "w", stdout)
#define mes(x, m) memset(x, m, sizeof(x))
#define Pii pair<int, int>
#define Pll pair<ll, ll>
#define INF 1e9+7
#define Pi 4.0*atan(1.0)

#define lowbit(x) (x&(-x))
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

typedef long long ll;
typedef unsigned long long ull;
const double eps = 1e-12;
const int maxn = 35;
using namespace std;

inline int read(){
    int x(0),op(1);
    char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')op=-1,ch=getchar();}
    while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
    return x*op;
}
inline ll extend_gcd(ll a,ll b,ll &x,ll &y)
{
    if(0==b){
        x=1,y=0;
        return a;
    }
    else{
        ll d=extend_gcd(b,a%b,x,y);
        ll tmp=x;
        x=y;
        y=tmp-a/b*x;
        return d;
    }
}
int main()
{
    ll a,b,c,k;
    while(~scanf("%lld%lld%lld%lld",&a,&b,&c,&k)){
        if(0==a&&0==b&&0==c&&0==k){
            break;
        }
        ll x,y;
        ll m=1LL<<k;
        b-=a;
        ll d=extend_gcd(c,m,x,y);
        if(b%d){
            printf("FOREVER\n");
            continue;
        }
        else{
            ll mod=m/d;
            ll res=(b/d*x%mod+mod)%mod;
            printf("%lld\n",res);
        }
    }
    return 0;
}