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## Discussion of Problem 1110. Power

This is the right solution
Posted by netman 24 Mar 2013 15:18
This is the right solution
Sorry for my english

var ans: array[1..1000] of longint;
n,m,y,q,i: longint;

function BinPow(x,y: longint): longint;
begin
if y=1 then BinPow:=x
else if y mod 2=0 then BinPow:=sqr(BinPow(x,y div 2)) mod M
else BinPow:=(sqr(BinPow(x,y div 2))*x) mod M;
end;

begin

q:=0;

for i:=0 to m-1 do
begin
if BinPow(i,n)=y then
begin
inc(q);
ans[q]:=i;
end;
end;

for i:=1 to q do
write(ans[i],' ');

if q=0 then writeln(-1);
end.
Re: This is the right solution
Posted by ELDVN 2 Nov 2015 00:11
My solve in C++:
#include <iostream>
#include <string>
#include <vector>
#include <set>
#include <queue>
#include <map>
#include <stack>
#include <algorithm>
#include <bitset>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <cstdio>
#include <iomanip>

#define F first
#define S second
#define ll long long
#define len length()
#define sqr(x) x*x
#define pb push_back
#define mp make_pair
#define sz(x) ((int) (x).size())
#define all(x) x.begin(), x.end()
#define allr(x) x.rbegin(), x.rend()
#define bp(x) __builtin_popcount(x)
#define INF numeric_limits<long long int>::max()
#define frp freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout);
#define forit(it, s) for(__typeof(s.begin()) it = s.begin(); it != s.end(); it++)

const int maxn = (int)1e6;
const int mod = (int)1e9 + 7;

using namespace std;

__int64 n,m,y;

int binpow(int a, int n){
if (n == 0)
return 1;
if (n % 2 == 1)
return (binpow (a, n-1)*a)%m;
else{
int b = (binpow(a, n/2));
return (b*b)%m;
}
}

bool ok;

main(){
scanf("%I64d%I64d%I64d",&n,&m,&y);
for(int i=0; i < m; i++){
if(binpow(i,n)%m==y){
cout<<i<<" ";
ok=1;
}
}
if(!ok)
return cout<<"-1",0;

return 0;
}
Re: This is the right solution
Posted by RobinGud 6 Feb 2017 02:07
What it is?
else if y mod 2=0 then BinPow:=sqr(BinPow(x,y div 2)) mod M
else BinPow:=(sqr(BinPow(x,y div 2))*x) mod M;