题目链接:http://poj.org/problem?id=2109
输入n,p,求一个k,是的k^n=p。
这道题目用数学公式推一推, logk(p)=n,换底公式一用,logp/logk=n--->logk = logp/n----->k=2^(logp*(1/n))----->k=p^(1/n)
然后用一下Pow函数即可,注意,这里的p很大,所以用double存(double能存接近10^308那么大的数字)。
#include<stdio.h>
#include<iostream>
#include<string>
#include<string.h>
#include<algorithm>
#include<iomanip>
#include<vector>
#include<time.h>
#include<queue>
#include<stack>
#include<iterator>
#include<math.h>
#include<stdlib.h>
#include<limits.h>
#include<map>
#include<set>
//#define ONLINE_JUDGE
#define eps 1e-8
#define INF 0x7fffffff
#define FOR(i,a) for((i)=0;i<(a);(i)++)
#define MEM(a) (memset((a),0,sizeof(a)))
#define sfs(a) scanf("%s",a)
#define sf(a) scanf("%d",&a)
#define sfI(a) scanf("%I64d",&a)
#define pf(a) printf("%d\n",a)
#define pfI(a) printf("%I64d\n",a)
#define pfs(a) printf("%s\n",a)
#define sfd(a,b) scanf("%d%d",&a,&b)
#define sft(a,b,c)scanf("%d%d%d",&a,&b,&c)
#define for1(i,a,b) for(int i=(a);i<b;i++)
#define for2(i,a,b) for(int i=(a);i<=b;i++)
#define for3(i,a,b)for(int i=(b);i>=a;i--)
#define MEM1(a) memset(a,0,sizeof(a))
#define MEM2(a) memset(a,-1,sizeof(a))
const double PI=acos(-1.0);
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
using namespace std;
#define ll __int64
int n,m,r;
#define N 110
int main(){
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
#endif
double n,p;
while(scanf("%lf%lf",&n,&p)!=EOF){
printf("%.0f\n",pow(p,1.0/n));
}
return 0;
}