zoukankan      html  css  js  c++  java
  • poj 2115 C Looooops 扩展欧几里德

    C Looooops
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 23616   Accepted: 6517

    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

    题意:(a+bx)%(2^k)==c,求x最小值;
    思路:解扩展欧几里德;
    #include<iostream>
    #include<string>
    #include<cstring>
    #include<algorithm>
    #include<cstdio>
    using namespace std;
    #define ll long long
    #define esp 1e-13
    const int N=1e4+10,M=1e6+50000,inf=1e9+10,mod=1000000007;
    void extend_Euclid(ll a, ll b, ll &x, ll &y)
    {
        if(b == 0)
        {
            x = 1;
            y = 0;
            return;
        }
        extend_Euclid(b, a % b, x, y);
        ll tmp = x;
        x = y;
        y = tmp - (a / b) * y;
    }
    ll gcd(ll a,ll b)
    {
        if(b==0)
            return a;
        return gcd(b,a%b);
    }
    ll pow1(ll x)
    {
        ll sum=1;
        for(ll i=0;i<x;i++)
        sum*=2;
        return sum;
    }
    int main()
    {
        ll x,y,i,z,t;
        while(~scanf("%lld%lld%lld%lld",&x,&y,&i,&t))
        {
            if(x==0&&y==0&&i==0&&t==0)
            break;
            ll m=pow1(t);
            ll c=((y-x)%m+m)%m;
            ll j,k;
            if(c%gcd(m,i)==0)
            {
                extend_Euclid(i,m,j,k);
                ll ans=j*(c/gcd(m,i));
                m=m/gcd(m,i);
                printf("%lld
    ",(ans%m+m)%m);
            }
            else
            printf("FOREVER
    ");
        }
        return 0;
    }
  • 相关阅读:
    css引入方式
    HTML标签
    动态导入模块impoerlib
    pymysql连接数据库
    创建数据库表之引擎
    IO多路复用互动聊天,select函数监听
    欧拉筛法求素数个数
    与三角形相关的问题 WITH 有向面积
    时间复杂度的计算
    折半查找
  • 原文地址:https://www.cnblogs.com/jhz033/p/5766058.html
Copyright © 2011-2022 走看看