一本通 1615：【例 1】序列的第 k 个数

序列的第k个数

快速幂

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Code:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
//Mystery_Sky
//
#define Mod 200907
#define M 10001000
#define ll long long
int n, k, a[M];

ll quickPow(ll x, ll k)
{
	ll ret = 1;
	while(k) {
		if(k & 1) ret = (ret * x) % Mod;
		k >>= 1;
		x = (x * x) % Mod;
	}
	return ret % Mod;
}
int t;

int main() {
	scanf("%d", &t);
	while(t--) {
		ll a, b, c, k;
		scanf("%lld%lld%lld%lld", &a, &b, &c, &k);
		if((a + c) == (2 * b)) {
			ll d = (b - a) % Mod;
			printf("%lld
", (a + ((k - 1) * d) % Mod) % Mod);
		}
		else {
			ll q = (b / a) % Mod;
			printf("%lld
", (a % Mod * quickPow(q, k-1)) % Mod);
		}
	}
	return 0;
}