与502. 序列的第 k 个数一模一样

AC CODE

#include<queue>
#include<math.h>
#include<stdio.h>
#include<iostream>
#include<vector>
#include<iomanip>
#include<string.h>
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<utility>
#include<cstring>
#include<stack>
#include<fstream>
#include<string>
using namespace std;
#define LL long long
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const long long M = 200907;
long long q( long long a , long long b )
{
	long long ret = 1;
	while ( b != 0 )
	{
		if ( b % 2 == 1 ) 
		{
			ret = ( ret * a % M ) % M;
		}
		a = a * a % M;
		b /= 2;
	}
	return ret % M;
}
long long a , b , c;
long long k;
long long t;
int main()
{
	cin >> t;
	while ( t-- )
	{
		cin >> a >> b >> c >> k;
		if ( b - a == c - b ) 
		{
			long long d = ( b - a ) % M;
			cout << ( ( k - 1 ) % M * d + a % M ) % M << endl;
		}
		else 
		{
			long long d = b / a % M;
			cout << ( q( d , k -1 ) * a % M ) % M << endl;
		}
	}
	return 0;
}

#include<bits/stdc++.h>
using namespace std;
long long fp(long long a, long long b, long long m){
	long long res=1;
	a=a%m;
	while(b>0){
		if(b%2==1){
			res=(res*a)%m;
		}
		a=(a*a)%m;
		b/=2;
	}
	return res;
}
long long gt(long long a,long long b,long long c,long long k){
	if(b-a==c-b){
		long long d=b-a;
		return (a+(k-1)*d)%200907;
	}else{
		long long q=b/a;
		return (a*fp(q,k-1,200907))%200907;
	}
}
int main(){
	int t;
	cin >> t;
	while(t--){
		long long a,b,c,k;
		cin>>a>>b>>c>>k;
		long long rst=gt(a,b,c,k);
		cout<<rst<<endl;
	}
	return 0;
}