北邮OJ 980. 16校赛-R_clover's Challenge

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题目描述

    R_clover wants to challenge Mengmengda_wsw's math，so he give her a function f(x,k).
    f(x,k)=g8(x,k)
    gn(x,k)=gn−1(g(x,k),k)
    where x∈[0,230−1],k∈[0,215−1].
    g(x,k)=(R(x)<<15)+(L(x)⊕((R(x)∗k%(215−1)))
    R(x)=x&(215−1)
    L(x)=x>>15
    Here, ⊕ means xor， & means bitwise AND operation bit-by-bit.
    And f(x,k)∈[0,230−1]。

The g(x,k) is defined as followed:

    R_clover gives Mengmengda_wsw x and f(x,k), and lets her calculate the value of k.
    However, it is so easy for Mengmengda_wsw. She enumerate all kinds of k，which adds up to 215=32768，and she can calculate it using computer。
    But R_clover won't let she solve the problem with computer, so he increases the difficluty.
    He provides x,f(f(x,k1),k2) and Mengmengda_wsw has to calculate k1,k2.
    Now it is too hard for Mengmengda_wsw, can you help her?

输入格式

The first line contains an integer T, (1≤T≤40)，which indicates the number of test cases.
For the next T lines，each line contains two integers a,b where a,b∈[0,230−1]， respectively indicating x and f(f(x,k1),k2)(k1,k2∈[0,215−1]).

输出格式

For each case, print two integers in a line, respectively indicating k1,k2. It is guaranteed that for each case there is only one solution.

输入样例

11063899164 441357888

输出样例

24533 30870

一道模拟+枚举题，首先我们可以看到f(x,k)里k的范围是[0,2^15-1]，3W多一点，

其次，我们看到f(x)和g(x)的运算已经给出，我们可以分别写出逆函数_f(x)和_g(x)

题目给出，f(f(x,k1),k2)=y，那么我们已知用逆函数_f(x)枚举k2，逆求出所有可能的f(x,k1)的值，把k2和逆求出的f(x,k1)存在一个set里。

然后我们根据已知x的值，枚举k1，求出所有可能的f(x,k1)，如果跟之前求出的set里的f(x,k1)的值匹配上了，那么就可以得到k1和k2了。

#include<cstdio>#include<cmath>#include<algorithm>#include<map>using namespace std;int M=(1<<15)-1;int gx(int x,int k){    int tmp=x&(M);    int tmp2=x>>15;    return (tmp<<15) + ( tmp2^((tmp*k)%M) );}int fx(int x,int k){    for(int i=0;i<8;i++){        x=gx(x,k);    }    return x;} int gx2(int x,int k){    int tmp=x&(M);    int tmp2=x>>15;    return tmp2 + ( (tmp^((tmp2*k)%M))<<15 );}int fx2(int x,int k){    for(int i=0;i<8;i++){        x=gx2(x,k);    }    return x;}map<int,int>rec;int main(){    int t;    int a,b,x,y;    for(scanf("%d",&t);t--;){        scanf("%d%d",&x,&y);        rec.clear();        for(int i=0;i<M;i++){            rec[fx(x,i)]=i;        }        for(int i=M;i>0;i--){            int t=fx2(y,i);            if(rec[t]){                printf("%d %d\n",rec[t],i);                break ;            }        }    }    return 0;}

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