C Looooops POJ - 2115

題目：

A Compiler Mystery: We are given a C-language style for loop of type

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

題意：

初始值為A，每次可以增加C，值要mod2^k,問mod后的值如果等于B，增加了幾次C，如果無法等于B輸出FOREVER

題解：

看一下我的推導：

你會發現其實就是擴展歐幾里得的模板題，并求出最小正整數解
我們可以用 (x0 % b1 + b1 ) % b1得到它的最小正整數解了
此處x0= x * c / gcd(a,b)
詳細證明看下面博客
擴展歐幾里得講解

代碼：

#include<iostream> #include<vector> #include<string> #include<cmath> #include<algorithm> #include<cstdio> #include<cstring> #include<list> using namespace std; typedef long long ll; ll exgcd(ll a,ll b,ll &x,ll &y)//擴展歐幾里得算法 {if(b==0){x=1;y=0;return a; //到達遞歸邊界開始向上一層返回}ll gcd=exgcd(b,a%b,x,y);ll y1=y; //把x y變成上一層的ll x1=x;y=x1-(a/b)*y1;x=y1;return gcd; //得到a b的最大公因數 }int main() {ll A,B,C,K;while(cin>>A>>B>>C>>K){if(A==0&&B==0&&C==0&&K==0)break;ll x,y;ll a=C;ll b=(ll)1<<K;ll c=B-A;ll gcd=exgcd(a,b,x,y);if(c%gcd!=0){cout<<"FOREVER"<<endl;}else {x=(x*(c/gcd))%b;x=(x%(b/gcd)+b/gcd)%(b/gcd);cout<<x<<endl;}}return 0; }

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