Codeforces 189A About Bacteria

Codeforces Round #125 (Div. 1) A

A. About Bacteria

time 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 ≤ 10⁶) — 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

没有计算最坏情况是否超过int64，导致wa了一次。

这题需要推导一下再做

题意是一个培养皿里面的细菌前一秒钟有x个，下一秒就会繁殖到kx+b个。

假设在第一次试验中，这个培养皿第一秒钟放进去1个，过了n秒钟达到了z个。

当第二次实验在培养皿第一秒钟放进去t个，问至少经过多少秒钟使得培养皿的细菌数大于等于z个。

要把递推式转化成闭形式（即得到函数F(n)表示n秒钟细菌有F(n)个）

代码如下

#include <stdio.h>

int main()
{
    int i,ans,j,k,b,n,t;
    __int64 l,r;
    scanf("%d%d%d%d",&k,&b,&n,&t);
    if (k==1)
    {
        l=n;
        l=1+l*b-t;
        printf("%I64d\n",l/b+(l%b!=0)>=0?l/b+(l%b!=0):0);
        return 0;
    }
    r=k;
    r=r*t+b-t;
    l=k+b-1;
    ans=0;
    while(l<=r)
    {
        ans++;
        l*=k;
    }
    ans--;
    printf("%d\n",n-ans>=0?(n-ans):0);
    return 0;
}