sicily 1394 Root of the Problem

这道题理解了题意应该就很简单了吧..我用的是枚举，同时利用了指数函数的单调递增性，一旦试到某个A的N次方减去B大于当前的min值，就跳出循环，因为就算继续下去，只会得到更大的A^N -B。

 

 

// source code of submission 733659, Zhongshan University Online Judge System#include <iostream>#include <cmath>using namespace std;int main(){ int b,n,a,i,temp; while(cin>>b>>n) { int min = 1000000; if(b==0&&n==0) break; for(i = 1;;i++) { temp = 1; for(int k = 1;k<=n;k++) temp *= i; int dis = fabs(b - temp); if(min>dis) min = dis; else {i = i -1;break;} } cout<<i<<endl; } return 0;}