C Looooops
Time Limit:1000MSMemory Limit:65536K
Total Submissions:20515Accepted:5531
Description
A Compiler Mystery: We are given a C-language style for loop of typefor (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 2 163 7 2 167 3 2 163 4 2 160 0 0 0
Sample Output
0232766FOREVER
Source
题意：求for (int i=A; i!=B; i+=C) i%=2^k;循环了多少次。
分析：由题可以看出循环终止的时候i==B；此时有（A+Cx）%2^k==B；即：C*x – 2^k*y==B-A。标准欧几里德。
#include <iostream>#include <cstdio>#include <cstring>#include <stack>#include <queue>#include <map>#include <set>#include <vector>#include <cmath>#include <algorithm>using namespace std;const double eps = 1e-6;const double pi = acos(-1.0);const int INF = 0x3f3f3f3f;const int MOD = 1000000007;#define ll long long#define CL(a) memset(a,0,sizeof(a))void exgcd(ll a, ll b, ll &d, ll &x, ll &y){if(!b){d=a; x=1; y=0;}else{exgcd(b, a%b, d, y, x);y-=x*(a/b);}}int main (){ll A,B,C,k;while (scanf("%lld%lld%lld%lld",&A,&B,&C,&k),(A+B+C+k)){ll a,b,d,x,y;ll c=B-A;if (!c) {cout<<"0"<<endl; continue;}a=C; b=(ll)1<<k;//位运算2^kexgcd(a, b, d, x, y);if (c%d) {cout<<"FOREVER"<<endl; continue;}ll dm=b/d;x = ((x*c/d)%dm+dm)%dm;cout<<x<<endl;}return 0;}
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，每一件事都要用多方面的角度来看它
