zoukankan      html  css  js  c++  java
  • C Looooops(poj2115+扩展欧几里德)

    C Looooops

    Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

    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 < 2 k) modulo 2 k.

    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 < 2 k) 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

    题意:for(i=A;i!=B;i+=C){i%(2^k)};问你循环执行几次?

    思路:先假设等式成立:(A+x*C)%(2^k)=B

    变形(2^k)*y+B=A+C*x ==> C*x+(-(2^k)*y)=B-A;

                                                       ax+by=c

    所以现在你知道怎么做了吧。哈哈!

    转载请注明出处:http://www.cnblogs.com/yuyixingkong/

    题目链接:http://poj.org/problem?id=2115

    #include<stdio.h>
    #define LL unsigned long long
    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("%llu%llu%llu%llu",&A,&B,&C,&k),(A+B+C+k))
        {
            LL a,b,c,d,x,y,dm;
            c=B-A;
            if(c==0){printf("0
    ");continue;}
            a=C;
            b=(LL)1<<k;
            exgcd(a,b,d,x,y);
            if(c%d){ printf("FOREVER
    ");continue;}
            dm=b/d;
            x=(((x*c/d)%dm)+dm)%dm;
    
            printf("%llu
    ",x);
        }
        return 0;
    }
  • 相关阅读:
    Azure WAF防火墙工作原理分析和配置向导
    多云时代,海外微软Azure云与国内阿里云专线打通性能测试
    【Hololens】微软Hololens虚拟现实视频集
    【Azure】Azure学习方法和学习资料
    LINUX ON AZURE 安全建议(全)
    编程行业之网络贩卖生存
    Bootstrap
    我谷首开博客
    Noip2018普及组初赛试题解题报告
    奇yin技巧
  • 原文地址:https://www.cnblogs.com/yuyixingkong/p/4472312.html
Copyright © 2011-2022 走看看