HDU 1063 Exponentiation(大数乘法)
Exponentiation
Time Limit: 2000/500 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Problem 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.
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
题意：给你R和n，求
解题思路：其实这是一个模拟大数乘法的过程，可以转化成n个R相乘，而实数R可以转化成整数p，等计算出最后结果之后再添小数点
小数点后的位数计算方法：假设一开始R小数点后的位数为k位，那么计算结果小数点后应有k*n位，暂时记为s位，那么①如果的位数<s的话，我们就需要在前添小数点再补0，如样例2；如果添了小数点后，小数点之后的位数末尾有0，则需要去掉末尾的0，如样例6
放上一些样例以供参考：
Input95.123 12
Output
548815620517731830194541.899025343415715973535967221869852721
Input
0.4321 20
Output
.00000005148554641076956121994511276767154838481760200726351203835429763013462401
Input
5.1234 15
Output
43992025569.928573701266488041146654993318703707511666295476720493953024
Input
6.7592 9
Output
29448126.764121021618164430206909037173276672
Input
98.999 10
Output
90429072743629540498.107596019456651774561044010001
Input
1.0100 12
Output
1.126825030131969720661201
Input
.00001 1
Output
.00001
Input
.12345 1
Output
.12345
Input
0001.1 1
Output
1.1
Input
1.1000 1
Output
1.1
Input
10.000 1
Output
10
Input
000.10 1
Output
.1
Input
000000 1
Output
0
Input
000.00 1
Output
0
Input
.00000 0
Output
0
Input
000010 1
Output
10
Input
000.10 1
Output
Input
0000.1 1
Output
.1
Input
00.111 1
Output
.111
Input
0.0001 1
Output
.0001
Input
0.0001 3
Output
.000000000001
Input
0.0010 1
Output
.001
Input
0.0010 3
Output
.000000001
Input
0.0100 1
Output
.01
Input
0.0100 3
Output
.000001
Input
0.1000 1
Output
.1
Input
0.1000 3
Output
.001
Input
1.0000 1
Output
1
Input
1.0000 3
Output
1
Input
1.0001 1
Output
1.0001
Input
1.0001 3
Output
1.003003001
Input
1.0010 1
Output
1.001
Input
1.0010 3
Output
1.003003001
Input
1.0100 1
Output
1.01
Input
1.0100 3
Output
1.030301
Input
1.1000 1
Output
1.1
Input
1.1000 3
Output
1.331
Input
10.000 1
Output
10
Input
10.000 3
Output
1000
Input
10.001 1
Output
10.001
Input
10.001 3
Output
1000.300030001
Input
10.010 1
Output
10.01
Input
10.010 3
Output
1003.003001
Input
10.100 1
Output
10.1
Input
10.100 3
Output
1030.301
Input
99.000 1
Output
99
Input
99.000 3
Output
970299
Input
99.001 1
Output
99.001
Input
99.001 3
Output
970328.403297001
Input
99.010 1
Output
99.01
Input
99.010 3
Output
970593.059701
Input
99.100 1
Output
99.1
Input
99.100 3
Output
973242.271
Input
99.998 1
Output
99.998
Input
99.998 3
Output
999940.001199992
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<stack>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<stdlib.h>
#include<cmath>
#include<string>
#include<algorithm>
#include<iostream>
#define exp 1e-10
using namespace std;
const int N = 200;
const int inf = 2000000001;
const int mod = 2009;
char s[11];
int a[N];
int main()
{
int n,i,j,k,p,l;
while(~scanf("%s%d",s,&n))
{
k=p=0;
memset(a,0,sizeof(a));
for(i=j=0;s[i]!='\0';i++)
if(s[i]!='.')
p=p*10+s[i]-'0';
else
k=strlen(s)-i-1;
k*=n;
if(p==0)
{
puts("0");
continue;
}
a[j++]=1;
for(i=0;i<n;i++)
{
for(l=0;l<j;l++)
a[l]=a[l]*p;
l=0;
while(a[l]>0||l<j)
{
a[l+1]+=a[l]/10;
a[l]%=10;
l++;
}
j=l;
}
if(k>j)
{
printf(".");
for(i=0;i<k-j;i++)
printf("0");
}
for(l=0;l<k&&!a[l];l++);
for(i=j-1;i>=l;i--)
{
if(i+1==k)
printf(".");
printf("%d",a[i]);
}
puts("");
}
return 0;
}
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