zoukankan      html  css  js  c++  java
  • poj2115 C Looooops

    C Looooops
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 29262   Accepted: 8441

    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

    大致题意:求在mod 2^k下的一个for循环要运行多少次.
    分析:列出同余式,扩展欧几里得解决.
    #include <cstdio>
    #include <cstring>
    #include <iostream>
    #include <algorithm>
    
    using namespace std;
    
    typedef long long ll;
    ll a,b,c,k,A,B,C,d;
    
    ll qpow(ll b)
    {
        ll res = 1,x = 2;
        while (b)
        {
            if (b & 1)
                res *= x;
            x *= x;
            b >>= 1;
        }
        return res;
    }
    
    ll gcd(ll a,ll b)
    {
        if (!b)
            return a;
        return gcd(b,a % b);
    }
    
    void exgcd(ll a,ll b,ll &x,ll &y)
    {
        if (!b)
        {
            x = 1;
            y = 0;
            return;
        }
        exgcd(b,a % b,x,y);
        ll t = x;
        x = y;
        y = t - (a / b) * y;
    }
    
    int main()
    {
        while (scanf("%lld%lld%lld%lld",&a,&b,&c,&k) == 4)
        {
            if (!a && !b && !c && !k)
                break;
            C = b - a;
            A = c;
            B = qpow(k);
            d = gcd(A,B);
            if (C % d != 0 || (b >= B || a >= B || c >= B))
               printf("FOREVER
    ");
            else
            {
                ll x,y;
                exgcd(A,B,x,y);
                x = x * C / d;
                B /= d;
                x = (x % B + B) % B;
                printf("%lld
    ",x);
            }
        }
    
        return 0;
    }
  • 相关阅读:
    《构建之法》阅读报告
    教务管理系统类图及数据库E/R图
    设计模式:抽象工厂
    结对项目:四则运算程序测试
    Leetcode笔记之57和为s的连续正数序列
    Leetcode笔记之1103分糖果 II
    Leetcode笔记之199二叉树的右视图
    每日Scrum(9)
    每日Scrum(7)
    每日Scrum(6)
  • 原文地址:https://www.cnblogs.com/zbtrs/p/8073751.html
Copyright © 2011-2022 走看看