标签：$2 input follow alc 要求 lan each end https
https://cn.vjudge.net/problem/POJ-2115
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 $2^k$.
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.
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.
3 3 2 16 3 7 2 16 7 3 2 16 3 4 2 16 0 0 0 0
0 2 32766 FOREVER
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;
}
标签：$2 input follow alc 要求 lan each end https
原文地址：https://www.cnblogs.com/sahdsg/p/10817208.html
