時間限制?2000 ms?內存限制?65536 KB

題目描述

??? 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?

輸入格式

?? ?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]).

輸出格式

?? ?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.

輸入樣例

1 1063899164 441357888

輸出樣例

24533 30870 一道模擬+枚舉題，首先我們可以看到f(x,k)里k的范圍是[0,2^15-1]，3W多一點，

其次，我們看到f(x)和g(x)的運算已經給出，我們可以分別寫出逆函數_f(x)和_g(x)

題目給出，f(f(x,k1),k2)=y，那么我們已知用逆函數_f(x)枚舉k2，逆求出所有可能的f(x,k1)的值，把k2和逆求出的f(x,k1)存在一個set里。

然后我們根據已知x的值，枚舉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; }

總結

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