Codeforces Round #125 (Div. 2) / C. About Bacteria

C. About Bacteriatime limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Qwerty the Ranger took up a government job and arrived on planet Mars. He should stay in the secret lab and conduct some experiments on bacteria that have funny and abnormal properties. The job isn't difficult, but the salary is high.

At the beginning of the first experiment there is a single bacterium in the test tube. Every second each bacterium in the test tube divides itself into k bacteria.
After that some abnormal effects create b more bacteria in the test tube. Thus, if at the beginning of some second the test tube had x bacteria,
then at the end of the second it will have kx + b bacteria.

The experiment showed that after n seconds there were exactly z bacteria
and the experiment ended at this point.

For the second experiment Qwerty is going to sterilize the test tube and put there t bacteria. He hasn't started the experiment yet but he already wonders,
how many seconds he will need to grow at least z bacteria. The ranger thinks that the bacteria will divide by the same rule as in the first experiment.

Help Qwerty and find the minimum number of seconds needed to get a tube with at least z bacteria in the second experiment.

Input

The first line contains four space-separated integers k, b, n and t (1 ≤ k, b, n, t ≤ 10^{6) —
the parameters of bacterial growth, the time Qwerty needed to grow z bacteria in the first experiment and the initial number of bacteria in the second experiment,
correspondingly.}

Output

Print a single number — the minimum number of seconds Qwerty needs to grow at least z bacteria in the tube.

Sample test(s)input

3 1 3 5

output

2

input

1 4 4 7

output

3

input

2 2 4 100

output

0

很簡單的一道數學題，腦殘當時沒想出來。。。。。。

題目說1個細菌按公式num=k*num+b繁殖到z個需要n秒。然後問現在裡面有t個，則繁殖到z個需要幾秒。。。

剛開始在想z沒有給出還得求出來，但是數又很大沒法做。。。

但其實z根本不用算。。。

只要找到一個num剛好不大於t時（再繁殖一次就超過t了），則所求時間（t繁殖到z的時間）一定就是從num繁殖到z的時間。。。。。。

#include <iostream>using namespace std;int main(void){    long long k, b, n, t;    cin >> k >> b >> n >> t;    long long curr = 1, time = 0;    while (curr < t) {        curr = k * curr + b;        ++time;    }    if (curr > t) --time;    if (time > n) time = n;    cout << n - time<< endl;    return 0;}