Description
Current work in cryptography involves (among other things) large prime numbers and computing powers of numbers among these primes. Work in this area has resulted in the practical use of results from number theory and other branches of mathematics once considered to be only of theoretical interest.
This problem involves the efficient computation of integer roots of numbers.
Given an integer n>=1 and an integer p>= 1 you have to write a program that determines the n th positive root of p. In this problem, given such integers n and p, p will always be of the form k to the nth. power, for an integer k (this integer is what your program must find).
This problem involves the efficient computation of integer roots of numbers.
Given an integer n>=1 and an integer p>= 1 you have to write a program that determines the n th positive root of p. In this problem, given such integers n and p, p will always be of the form k to the nth. power, for an integer k (this integer is what your program must find).
Input
The input consists of a sequence of integer pairs n and p with each integer on a line by itself. For all such pairs 1<=n<= 200, 1<=p<10101 and there exists an integer k, 1<=k<=109 such that kn = p.
Output
For each integer pair n and p the value k should be printed, i.e., the number k such that k n =p.
Sample Input
2 16
3 27
7 4357186184021382204544
Sample Output
4
3
1234
poj最近在维护……然后一段时间内就都不能提交了。。
先挂着。。
方法比较简单，考虑到结果的大小是有二分性的，
所以可以二分一个值，然后判断是否合法即可。
写个结构体，然后就简单了很多……
说说另外的做法。。
本来看到求x^m=a的x，想到牛顿迭代就来看看……结果还是高精度。。
还是提一提吧。。
我们要求解x^m=a的解x，设f(x)=x^m-a，那么题目转化为求f(x)的零点。
写出f(x)的近似泰勒展开，，（泰勒级数不会自行百度= =）
f(x)=f(x0)+(x-x0)*f'(x0)=0
其中f'(x0)是f(x0)的一阶导数，x0是一个近似解，
那么解出x就可以得到一个方程的近似解。。
注意是近似的，因为原本的展开是无穷的，这只是一个近似展开。。
但是显然精度很大时候会不够……
这个时候就可以利用一个迭代的思想：
上面的式子=0两边同时除以f'(x0)，得到：
x=x0-f(x0)/f'(x0)
用x作为新的x0继续迭代，直到x在某一个精度范围内，
那么就得到了这个方程的某个近似解。
对于这题，f(x)=x^m-a，
f'(x)=m*x^(m-1)
根据这个迭代即可。
但是事实上如果要高精度的话，这个迭代会牵扯到高精度除法……
解决方法挺简单，解除f(x)=(1/x)^m-a的零点（模拟小数高精度），
然后就只用做一遍高精度除法了。
总的来说还是二分简便……但是乘法一慢就gg了= =
当然FFT什么的当然好哇。。
等到poj恢复了再交一发，现在精神AC一下……=v=
#include<bits/stdc++.h>
using namespace std;
const int
maxn=505;
struct BigNum{
int len,a[maxn];
BigNum(){len=0;memset(a,0,sizeof(a));}
void read(){
char t[maxn];
scanf("%s",t+1);
len=strlen(t+1);
for (int i=1;i<=len;i++) a[i]=t[len-i+1]-'0';
}
void print(){
for (int i=len;i;i--) printf("%d",a[i]);
putchar('\n');
}
BigNum operator =(int x){
char t[maxn];
sprintf(t+1,"%d",x);
len=strlen(t+1);
for (int i=1;i<=len;i++) a[i]=t[len-i+1]-'0';
return (*this);
}
BigNum operator =(BigNum x){
len=x.len;
for (int i=1;i<=len;i++) a[i]=x.a[i];
return (*this);
}
bool operator <(BigNum x){
if (len<x.len) return 1;
if (len>x.len) return 0;
for (int i=len;i;i--){
if (a[i]<x.a[i]) return 1;
if (a[i]>x.a[i]) return 0;
}
return 0;
}
bool operator >(BigNum x){return x<(*this);}
bool operator >=(BigNum x){return !((*this)<x);}
bool operator <=(BigNum x){return !((*this)>x);}
bool operator ==(BigNum x){return (*this)<=x && (*this)>=x;}
BigNum operator +(BigNum x){
BigNum y;
y.len=max(len,x.len);
for (int i=1;i<=y.len;i++){
y.a[i]+=x.a[i]+a[i];
y.a[i+1]=y.a[i]/10;
y.a[i]%=10;
}
if (y.a[y.len+1]) y.len++;
return y;
}
BigNum operator -(BigNum x){
BigNum mi,ma;
if ((*this)>x) ma=(*this),mi=x;
else ma=x,mi=(*this);
for (int i=1;i<=ma.len;i++){
if (ma.a[i]-mi.a[i]>=0) ma.a[i]=ma.a[i]-mi.a[i];
else ma.a[i]=mi.a[i]-ma.a[i],ma.a[i+1]--;
}
return ma;
}
BigNum operator +(int x){BigNum y;y=x;return (*this)+y;}
BigNum operator -(int x){BigNum y;y=x;return (*this)-y;}
BigNum operator *(BigNum x){
BigNum y;
y.len=x.len+len-1;
for (int i=1;i<=len;i++)
for (int j=1;j<=x.len;j++){
y.a[i+j-1]+=x.a[j]*a[i];
y.a[i+j]+=y.a[i+j-1]/10;
y.a[i+j-1]%=10;
}
if (y.a[y.len+1]) y.len++;
return y;
}
BigNum operator /(int x){
BigNum y;int t=0;
y.len=len;
while (t<x) t=t*10+a[y.len--];
y.a[++y.len]=t/x;
for (int j=y.len-1;j;j--){
t=t%x*10+a[j];
y.a[j]=t/x;
}
return y;
}
BigNum operator ^(int y){
BigNum z,e;
z=1;e=(*this);
while (y){
if (y&1) z=z*e;
e=e*e;y>>=1;
}
return z;
}
};
int main(){
BigNum a,l,r,mid,tmp;
int b;
while (~scanf("%d",&b)){
a.read();
l=1;r=a/(b-1);
while (l<=r){
mid=(l+r)/2;
tmp=mid^b;
if (tmp==a){mid.print();break;}
if (tmp>a) r=mid-1;
else l=mid+1;
}
}
return 0;
}