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
大意：培养细菌，1个细菌每秒增张为k*x+b,求t个细菌经过多少秒后增长到Z个。n是1个细菌经过n秒增加到Z，t是第二组刚开始有t个细菌
由于——int64的范围为9*10^19左右，但是此题如果直接做t*(k*x+b)，很有可能超__int64
#include<iostream>
#include<cstring>
#include<cstdio>
#include<string>
#include<cmath>
#include<algorithm>
#define LL __int64
#define inf 0x3f3f3f3f
using namespace std;
LL p[1000000];
int main()
{
LL n,m,i,j,k;
LL b,t;
while(~scanf("%I64d %I64d %I64d%I64d",&k,&b,&n,&t))
{
LL tmp=0;
LL x=1;
while(tmp<=n&&t>=x)
{
x=k*x+b;
tmp++;
}
printf("%I64d\n",n-tmp+1);//用n-(1个细菌增加到t个所用的时间)=t增加到Z所用的时间<span id="transmark"></span>
}
return 0;
}
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