序列的第k个数

快速幂

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\n", (a + ((k - 1) * d) % Mod) % Mod);
        }
        else {
            ll q = (b / a) % Mod;
            printf("%lld\n", (a % Mod * quickPow(q, k-1)) % Mod);
        }
    }
    return 0;
}

一本通 1615：【例 1】序列的第 k 个数

原文：https://www.cnblogs.com/Benjamin-cpp/p/11122817.html