C Looooops
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
题目类型：一元一次同余方程
算法分析：题目要求解的是C * x = (B – A) mod (1 << k)的解x，直接使用扩展欧几里得算法求解即可，注意要强制将(1 << k)转换成long long型，否则会WA
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#include <set> #include <bitset> #include <list> #include <map> #include <stack> #include <queue> #include <deque> #include <string> #include <vector> #include <ios> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <algorithm> #include <utility> #include <complex> #include <numeric> #include <functional> #include <cmath> #include <ctime> #include <climits> #include <cstdarg> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <cassert> #define lson rt << 1, l, m #define rson rt << 1 | 1, m + 1, r using namespace std; const int INF = 0x7FFFFFFF; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int MOD = 7; const int maxn = 666; long long gcd_ex (long long a, long long b, long long &x, long long &y) { if (b == 0) { x = 1; y = 0; return a; } long long d = gcd_ex (b, a % b, y, x); y = y - a / b * x; return d; } long long mod_equ (long long a, long long b, long long c) { long long x, y; long long d = gcd_ex (a, b, x, y); if (c % d) return -1; x = (x * c / d) % (b / d); return (x + b / d) % (b / d); } int main() { // ifstream cin ("aaa.txt"); long long A, B, C, k; while (cin >> A >> B >> C >> k) { if (A == 0 && B == 0 && C == 0 && k == 0) break; long long ans = mod_equ (C, (long long) 1 << k, B - A); if (ans == -1) cout << "FOREVER" << endl; else cout << ans << endl; } return 0; } |
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