zoukankan      html  css  js  c++  java
  • POJ 2115 C Looooops

    https://cn.vjudge.net/problem/POJ-2115

    题目

    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 $0 leqslant x < 2^k$) 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
    

    题解

    写出线性同余方程

    [A+Cxequiv B pmod {2^k}]

    整理得

    [Cxequiv B-A pmod {2^k}]

    有解的充要条件是$ ext{GCD}(C,2^k)mid{B-A}$

    两边同除$ ext{GCD}(C,2^k)$,得

    [frac{C}{g} xequiv frac{B-A}{g} pmod {frac{2^k}{g}}]

    然后就可以两边同乘以逆元了,得到模$frac{2^k}{g}$下的唯一解,则模$2^k$下有$g$个解= =

    但是题目只要求最小的……

    还有一个坑是爆int,在移位的时候一定要用1LL

    真的服了,起名llabs还能撞名字= =

    AC代码

    #include<cstdio>
    #include<cstdlib>
    #include<cctype>
    #include<cstring>
    #include<algorithm>
    #include<set>
    #define REP(r,x,y) for(register int r=(x); r<(y); r++)
    #define REPE(r,x,y) for(register int r=(x); r<=(y); r++)
    #define PERE(r,x,y) for(register int r=(x); r>=(y); r--)
    #ifdef sahdsg
    #define DBG(...) printf(__VA_ARGS__),fflush(stdout)
    #else
    #define DBG(...) (void)0
    #endif
    using namespace std;
    typedef long long LL;
    typedef unsigned long long ULL;
    char ch; int si;
    
    #define gc() getchar()
    
    template<class T>
    inline void read(T &x) {
        x=0; si=1; for(ch=gc();!isdigit(ch) && ch!='-';ch=gc());
        if(ch=='-'){si=-1,ch=gc();} for(;isdigit(ch);ch=gc())x=x*10+ch-'0';
        x*=si;
    }
    template<class T, class...A> inline void read(T &x, A&...a){read(x); read(a...);}
    
    inline LL qmul(LL a, LL b, LL p) {
        LL ans=0;
        for(;b;b>>=1) {
            if(b&1) ans=(ans+a)%p;
            a=(a*2)%p;
        }
        return ans;
    }
    
    inline LL GCD(LL a, LL b) {
        return b==0? a:GCD(b,a%b);
    }
    
    inline void exgcd(LL a, LL b, LL &x, LL &y) {
        if(b==0) {x=1, y=0; return;}
        exgcd(b,a%b,y,x);
        y-=a/b*x;
        //return d;
    }
    //inline LL llabs(LL x) {
    //    return x<0?-x:x;
    //}
    
    int main() {
    #ifdef sahdsg
        freopen("in.txt","r",stdin);
    #endif
        LL a,b,c,d;
        while(~scanf("%lld%lld%lld%lld", &a, &b, &c, &d) && (a||b||c||d)) {
            d=1LL<<d;
            //ct===b-a mod 1<<k
            b-=a;
            LL g=(GCD(c,d)); DBG("#%lld %lld %lld
    ",g, c, d);
            if(b%g) {
                puts("FOREVER");
                continue;
            }
            b/=g, c/=g, d/=g;
            LL x,t;
            exgcd(c,d,x,t);
            b*=x;
            b%=d; if(b<0) b+=d;
            printf("%lld
    ", b);
        }
        return 0;
    }
    
  • 相关阅读:
    架构师速成-怎样高效编程
    【论文笔记】Leveraging Datasets with Varying Annotations for Face Alignment via Deep Regression Network
    Redis数据类型--List
    python命令行传入参数
    python 连接ORacle11g
    sqlserver2016 kb补丁
    linux cat 文件编码
    python gtk 环境
    openstack kvm cannot set up guest memory 'pc.ram': Cannot allocate memory
    Mysql Explain 详解(转)
  • 原文地址:https://www.cnblogs.com/sahdsg/p/10817208.html
Copyright © 2011-2022 走看看