POJ2115 C looooops "Solving linear congruence equation"

Topic Links:

http://poj.org/problem?id=2115

Main topic:

For loop statements:

for (int i = A; I! = B; i + = C)

Statement 1;

It is known that I, A, B, and C are all K-binary unsigned integer types, giving values of a, B, C, K, and calculating and outputting statements 1

The number of executions, if infinite, then the direct output "FOREVER".

Ideas:

set the algorithm to perform X-steps, then the title becomes solution A + cx≡b (mod M) (m= 2^k). that is a + CX + my≡b.

CX + MY ≡b-a (M = 2^k) is changed in order to find linear congruence equation, simple linear congruence algorithm can be applied.

AC Code:

#include <iostream> #include <algorithm> #include <cstdio> #include <cstring>using namespace Std;__int64 a,b,c,k;__int64 a,b,c,d,x,y;void exgcd (__int64 a,__int64 b,__int64 &d,__int64 &x,__int64 &y) { C0/>if (!b)        x = 1, y = 0, d = A;    else    {        exgcd (b,a%b,d,y,x);        Y-= x * (A/b);    }} int main () {while    (cin >> A >> B >> C >> k && (a| | b| | c| | k))    {        a = C;        c = b-a;        b = (__int64) 1 << k;        EXGCD (a,b,d,x,y);        if (c% d! = 0)            cout << "FOREVER" << Endl;        else        {            __int64 ans,temp;            Ans = x * C/D;            temp = b/d;            ans = ans% temp + temp;            cout << ans% temp << endl;        }    }    return 0;}

POJ2115 C looooops "Solving linear congruence equation"