poj 1730 Perfect Pth Powers
2011-07-22 22:47
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#include<iostream> #include<math.h> using namespace std; #define eps 1e-2 //其他精度数也同样可以 int main() { int n; while(cin>>n&&n) { //n有负数的可能,比如 -1073741824,正解应该是 15, (-4)^15=-1073741824 int t=n>0?1:-1; for(int i=31;i>=1;--i) { double x=ceil(pow(double(n)*t,1.0/i))*t; //向上取整 double y=floor(pow(double(n)*t,1.0/i))*t; //向下取整 //这里如果改成double(n*t)或n*t*1.0或n*t+0.0都是 WA ,可能是在t=-1时,n*t会超int 的负数表示范围,所以在n*t前必须先将 n 转化成浮点数 if(fabs(pow(x+0.0,i+0.0)-n)<=eps||fabs(pow(y+0.0,i+0.0)-n)<=eps) { cout<<i<<endl; break; } } } return 0; }
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