C Looooops
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 29262 | Accepted: 8441 |
Description
A Compiler Mystery: We are given a C-language style for loop of type
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.
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.
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
大致题意:求在mod 2^k下的一个for循环要运行多少次.
分析:列出同余式,扩展欧几里得解决.
#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; typedef long long ll; ll a,b,c,k,A,B,C,d; ll qpow(ll b) { ll res = 1,x = 2; while (b) { if (b & 1) res *= x; x *= x; b >>= 1; } return res; } ll gcd(ll a,ll b) { if (!b) return a; return gcd(b,a % b); } void exgcd(ll a,ll b,ll &x,ll &y) { if (!b) { x = 1; y = 0; return; } exgcd(b,a % b,x,y); ll t = x; x = y; y = t - (a / b) * y; } int main() { while (scanf("%lld%lld%lld%lld",&a,&b,&c,&k) == 4) { if (!a && !b && !c && !k) break; C = b - a; A = c; B = qpow(k); d = gcd(A,B); if (C % d != 0 || (b >= B || a >= B || c >= B)) printf("FOREVER "); else { ll x,y; exgcd(A,B,x,y); x = x * C / d; B /= d; x = (x % B + B) % B; printf("%lld ",x); } } return 0; }