题意理解:
求解指数的Pow(A,N)的底A,使得pow(A,N)最接近B;
解题分析:
水题;
解题代码:
#include<iostream>
#include<math.h>
using namespace std;
int main(int argc, char *argv[]){
int B,N;
while(cin>>B>>N, B||N){
int preDiff=B;
int A;
int diff;
for(int i=1;i<=B;i++){
diff=pow(1.0*i,N)-B;
if(diff<0){
preDiff=abs(diff);
}else if(diff==0){
A=i;
break;
}else{
if(diff>=preDiff){
A=i-1;
}else{
A=i;
}
break;
}
}
cout<<A<<endl;
}
return 0;
}