# 北大poj-1001

Description

Problems involving the computation of exact values of very large magnitude and precision are common. For example, the computation of the national debt is a taxing experience for many computer systems.

This problem requires that you write a program to compute the exact value of R
n where R is a real number ( 0.0 < R < 99.999 ) and n is an integer such that 0 < n <= 25.

Input

The input will consist of a set of pairs of values for R and n. The R value will occupy columns 1 through 6, and the n value will be in columns 8 and 9.

Output

The output will consist of one line for each line of input giving the exact value of R^n. Leading zeros should be suppressed in the output. Insignificant trailing zeros must not be printed. Don’t print the decimal point if the result is an integer.

Sample Input

```95.123 12
0.4321 20
5.1234 15
6.7592  9
98.999 10
1.0100 12
```

Sample Output

```548815620517731830194541.899025343415715973535967221869852721
.00000005148554641076956121994511276767154838481760200726351203835429763013462401
43992025569.928573701266488041146654993318703707511666295476720493953024
29448126.764121021618164430206909037173276672
90429072743629540498.107596019456651774561044010001
1.126825030131969720661201```
```#include<stdio.h>
#include<string.h>

char *touzero(char *tmp)
{
if(*tmp=='0'&&*(tmp+1)!='\0')
{
tmp=tmp+1;
tmp=touzero(tmp);
}
return tmp;
}

int weizero(char *tmp,int m)
{
if(*tmp=='0')
{
m++;
*tmp='\0';
tmp=tmp-1;
m=weizero(tmp,m);
}
return m;
}

int main()
{
char s[7],d[150];
int e[150];
char *tmp;
int i,j,k,n,tmpint;
int m;
int c;

while(~scanf(" %s %d",s,&n))
{
c=0;
k=m=-1;
memset(e,0,sizeof(e));
memset(d,0,sizeof(d));
s[6]='\0';
tmp=s+5;
for(i=0;i<sizeof(s);i++)
{
if(s[i]=='.')
{
m=i;
k=4-weizero(tmp,k)-i;
break;
}
}
if(m!=-1)
{
for(i=m;i<(sizeof(s)-1);i++)
{
s[i]=s[i+1];
}
s[i]='\0';
}
tmp=s;
tmp=touzero(tmp);
strcpy(s,tmp);
i=0;
while(s[i]!='\0')i++;
j=i-1;
do
{
i--;
e[j-i]=(int)s[i]-48;
}while(i>0);
sscanf(s,"%d",&tmpint);
for(i=0;i<n-1;i++)
{
for(j=0;j<149;j++)
{
e[j]=e[j]*tmpint;
}
for(j=0;j<149;j++)
{
e[j+1]=e[j+1]+e[j]/10;
e[j]=e[j]%10;
}
}
tmpint=0;
m=0;
for(i=0;i<150;i++)
{
if(e[i]!=0){m=1;break;}
}
if(n==0)
printf("1\n");
else if(m==0)
printf("0\n");
else
{
for(i=0;i<150;i++)//输出
{
if(tmpint==0&&e[149-i]==0&&150-i>k*n)
{
continue;
}
tmpint=1;
if(i==150-k*n)
{
printf(".");
}
printf("%d",e[149-i]);
}
printf("\n");
}
}
return 0;
}
```

原文作者：Online Judge POJ
原文地址: https://www.cnblogs.com/bixiongquan/p/3204606.html
本文转自网络文章，转载此文章仅为分享知识，如有侵权，请联系博主进行删除。