POJ 2115 C Looooops

最新推荐文章于 2025-05-20 23:10:17 发布

banlun8031

最新推荐文章于 2025-05-20 23:10:17 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/sahdsg/p/10817208.html

版权

https://cn.vjudge.net/problem/POJ-2115

题目

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 $0 \leqslant x < 2^k$) 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

Sample Output

0
2
32766
FOREVER

题解

写出线性同余方程

\[A+Cx\equiv B \pmod {2^k}\]

整理得

\[Cx\equiv B-A \pmod {2^k}\]

有解的充要条件是$\text{GCD}(C,2^k)\mid{B-A}$

两边同除$\text{GCD}(C,2^k)$，得

\[\frac{C}{g} x\equiv \frac{B-A}{g} \pmod {\frac{2^k}{g}}\]

然后就可以两边同乘以逆元了，得到模$\frac{2^k}{g}$下的唯一解，则模$2^k$下有$g$个解= =

但是题目只要求最小的……

还有一个坑是爆int，在移位的时候一定要用1LL！

真的服了，起名llabs还能撞名字= =

AC代码

#include<cstdio>
#include<cstdlib>
#include<cctype>
#include<cstring>
#include<algorithm>
#include<set>
#define REP(r,x,y) for(register int r=(x); r<(y); r++)
#define REPE(r,x,y) for(register int r=(x); r<=(y); r++)
#define PERE(r,x,y) for(register int r=(x); r>=(y); r--)
#ifdef sahdsg
#define DBG(...) printf(__VA_ARGS__),fflush(stdout)
#else
#define DBG(...) (void)0
#endif
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
char ch; int si;

#define gc() getchar()

template<class T>
inline void read(T &x) {
    x=0; si=1; for(ch=gc();!isdigit(ch) && ch!='-';ch=gc());
    if(ch=='-'){si=-1,ch=gc();} for(;isdigit(ch);ch=gc())x=x*10+ch-'0';
    x*=si;
}
template<class T, class...A> inline void read(T &x, A&...a){read(x); read(a...);}

inline LL qmul(LL a, LL b, LL p) {
    LL ans=0;
    for(;b;b>>=1) {
        if(b&1) ans=(ans+a)%p;
        a=(a*2)%p;
    }
    return ans;
}

inline LL GCD(LL a, LL b) {
    return b==0? a:GCD(b,a%b);
}

inline void exgcd(LL a, LL b, LL &x, LL &y) {
    if(b==0) {x=1, y=0; return;}
    exgcd(b,a%b,y,x);
    y-=a/b*x;
    //return d;
}
//inline LL llabs(LL x) {
//    return x<0?-x:x;
//}

int main() {
#ifdef sahdsg
    freopen("in.txt","r",stdin);
#endif
    LL a,b,c,d;
    while(~scanf("%lld%lld%lld%lld", &a, &b, &c, &d) && (a||b||c||d)) {
        d=1LL<<d;
        //ct===b-a mod 1<<k
        b-=a;
        LL g=(GCD(c,d)); DBG("#%lld %lld %lld\n",g, c, d);
        if(b%g) {
            puts("FOREVER");
            continue;
        }
        b/=g, c/=g, d/=g;
        LL x,t;
        exgcd(c,d,x,t);
        b*=x;
        b%=d; if(b<0) b+=d;
        printf("%lld\n", b);
    }
    return 0;
}

转载于:https://www.cnblogs.com/sahdsg/p/10817208.html