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  • POJ2115 C Looooops (模线性方程)

    C Looooops
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 24380   Accepted: 6793

    Description

    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

     
    首先列出基本方程组 A+C*x=B+(2^k)*y   →   C*x+(2^k)*y=B-A  这样就满足了 ax+by=1 的模型,所以通过扩展欧几里得求出即可。
     
     1 #include <bits/stdc++.h>
     2 
     3 using namespace std;
     4 typedef long long LL;
     5 LL A,B,C,k;
     6 LL a,b,c,d,e,f,x,y;
     7 LL exgcd(LL a,LL b,LL &x,LL &y){
     8     if (b==0)
     9     {x=1;
    10      y=0;
    11      return a;
    12     }
    13     LL d=exgcd(b,a%b,x,y);
    14     LL t=x;
    15     x=y;
    16     y=t-(a/b)*y;
    17     return d;
    18 }
    19 int main(){
    20     freopen ("c.in","r",stdin);
    21     freopen ("c.out","w",stdout);
    22     int i,j;
    23     while (1)
    24     {scanf("%lld%lld%lld%lld",&A,&B,&C,&k);
    25      if (A==0 && B==0 && C==0 && k==0)
    26       break;
    27      a=C,b=1ll<<k,c=B-A;
    28      d=exgcd(a,b,x,y);
    29      if (c==0)
    30      {puts("0");
    31       continue;
    32      }
    33      if (c%d!=0)
    34      {puts("FOREVER");
    35       continue;
    36      }
    37      e=x*(c/d);
    38      f=(e%(b/d)+b/d)%(b/d);
    39      printf("%lld\n",f);
    40     }
    41     return 0;
    42 }
    未来是什么样,未来会发生什么,谁也不知道。 但是我知道, 起码从今天开始努力, 肯定比从明天开始努力, 要快一天实现梦想。 千里之行,始于足下! ——《那年那兔那些事儿》
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  • 原文地址:https://www.cnblogs.com/keximeiruguo/p/5990319.html
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