爱华网题目大意： K ^ N = P, 给N 和 P， 求K。数据规模：1<=n<= 200,1<=p<10101and there exists an integerk, 1<=k<=109。
思路：初看是大数运算，看了Disuss后（真不该……）发现竟然可以一句话AC……
| 12345678 | #include<stdio.h>#include<math.h>int main(){doublen, m;while(scanf("%lf%lf", &n,&m) != EOF)printf("%.0lfn",pow(m, 1 / n));} |
| 12345678 | #include<stdio.h>#include<math.h>int main(){doublen, m;while(scanf("%lf%lf", &n,&m) != EOF)printf("%.0lfn",pow(m, 1 / n));} |
#include<stdio.h>
#include<math.h>
int main()
{
double n,m;
while(scanf("%lf %lf", &n, &m) !=EOF)
printf("%.0lfn" ,pow(m, 1 / n));
}
类型长度(bit)有效数字绝对值范围
float326~710^(-37) ~ 10^38
double6415~1610^(-307) ~10^308
long double 12818~1910^(-4931) ~ 10 ^ 4932
double最大可存储10 ^ 308范围内的数，而p的范围才10 ^101，这很好理解。但double的精度才15-16位啊，存储之后精度损失这么严重，竟然有正确结果，这该怎么解释呢？
当然，pow(p ,1 / n)可以求K。受教了。
另外，可以用二分法AC此题，继续纠结精度问题……
?| 123456789101112131415161718192021222324252627282930313233 | #include<stdio.h>#include<math.h>#define eps 0.0000000001void init(),work();double n, m,k;int main(){init();return0;}void init(){while(scanf("%lf%lf", &n,&m) != EOF)work();}void work(){longlong left, right, mid;left= 0;right= 1000000002;while(left+ eps < right){mid= (left + right) / 2;if(pow(mid,n) - m > 0)right= mid;elseif(pow(mid,n) - m < 0)left= mid;else{printf("%.0ldn",mid) ;break;}}} |
#include<math.h>
#define eps 0.0000000001
void init(), work();
double n, m, k;
int main()
{
init();
return0;
}
void init()
{
while(scanf("%lf %lf", &n, &m) !=EOF)
work();
}
void work()
{
long longleft, right, mid;
left =0;
right =1000000002;
while(left +eps < right){
mid = (left + right) / 2;
if(pow(mid, n) - m > 0)
right = mid;
else
if(pow(mid, n) - m < 0)
left = mid;
else{
printf("%.0ldn", mid) ;
break;
}
}
}
