[Fast Power] [Extended Euclidean] [bsgs] [sdoi 2011] [bzoj 2242] Calculator

2242: [sdoi2011] Calculator

Time Limit: 10 Sec  Memory Limit: 512 MBSubmit: 2077  Solved: 812

Description

You are asked to design a calculator to complete the following three tasks:
1. Given Y, Z, P, calculate the value of y ^ Z mod P;
2. Given Y, Z, P, calculate the minimum non-negative integer that satisfies XY ≡ Z (mod P;
3. Given Y, Z, P, calculate the minimum non-negative integer that satisfies y ^ x ≡ Z (mod P.

Input

The input contains multiple groups of data.

The first line contains two positive integers t, K representing the number of data groups and the query type (for all data in a test point, the query type is the same ).
Each line below contains three positive integers Y, Z, and P, which describe a query.

Output

Output a line of answers for each query. For query Types 2 and 3, if the conditions are not met, the output is "orz, I cannot find X !", Note that there is a space between comma and I.

Sample Input

[Example input 1]

3 12 1 32 2 32 3 3

[Example input 2]

3 22 1 32 2 32 3 3

[Data scale and Conventions]

For 100% of data, 1 <= Y, Z, P <= 10 ^ 9, is the prime number, 1 <= T <= 10.

Sample output

[Example output 1]

212

[Sample output 2]

2
1
0

Question:

1. Fast Power.
2. Extend Euclidean.
3. bsgs, everyone should .. I'm not sure.
No, you can read it here --> it's good to write!

Code:

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<map>#include<algorithm>using namespace std;#define N 100100#define LL long longmap<LL,LL> hash;int in(){    int x=0; char ch=getchar();    while (ch<‘0‘ || ch>‘9‘) ch=getchar();    while (ch>=‘0‘ && ch<=‘9‘) x=x*10+ch-‘0‘,ch=getchar();    return x;}LL qp(LL x,LL y,LL p){    LL z=1;    while (y){        if (y&1) z=(z*x)%p;        x=(x*x)%p; y>>=1;    }    return z;}void work1(int y,int z,int p){    printf("%lld\n",qp(y,z,p));}LL gcd(LL x,LL y){    if (!y) return x;    return gcd(y,x%y);}void exgcd(LL a,LL b,LL &x,LL &y){    if (!b){        x=1,y=0;        return;    }    exgcd(b,a%b,x,y);    LL k=x; x=y;    y=k-a/b*y;}void work2(int y,int z,int p){    LL w=gcd(y,p);    if (z%w){        printf("Orz, I cannot find x!\n");        return;    }    LL x,k; exgcd(y,p,x,k);    x=x*z/w; x=(x+p)%p;    while (x<0) x+=p;    printf("%lld\n",x);}void work3(int y,int z,int p){    y%=p,z%=p; hash.clear();    if (!y && !z){        printf("1\n");        return;    }    if (!y){        printf("Orz, I cannot find x!\n");        return;    }    LL pp=ceil(sqrt(p)),v=qp(y,p-pp-1,p),k=1;    hash[1]=pp+1;    for (LL i=1; i<=pp; i++){        k=(k*y)%p;        if (!hash[k]) hash[k]=i;    }    LL ans=-1;    for (LL i=0; i<pp; i++){        LL j=hash[z];        if (j){            if (j==pp+1) j=0;            ans=(i*pp+j);            break;        }        z=(z*v)%p;    }    if (ans==-1) printf("Orz, I cannot find x!\n");    else printf("%lld\n",ans);}int main(){    int T=in(),k=in();    while (T--){        int y=in(),z=in(),p=in();        if (k==1) work1(y,z,p);        else if (k==2) work2(y,z,p);        else work3(y,z,p);    }    return 0;}

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[Fast Power] [Extended Euclidean] [bsgs] [sdoi 2011] [bzoj 2242] Calculator