分類開(kāi)篇語(yǔ): 第一個(gè)程序搞了好幾天，發(fā)現(xiàn)了很多問(wèn)題。POJ不保證按順序做且更新速度肯定不會(huì)很快。有些題自己做不出來(lái)借鑒別人的會(huì)注明出處。很多算法都需要從網(wǎng)上找，第一題的大浮點(diǎn)數(shù)相乘的核心算法就是這樣找來(lái)的。我心里明白，雖然AC了，但是邊緣數(shù)據(jù)處理的很粗糙，我自己都發(fā)現(xiàn)幾個(gè)bug了，但是依然AC了。
本題主要注意將字符串轉(zhuǎn)化為實(shí)際的數(shù)字然后借鑒數(shù)制的思想來(lái)進(jìn)行大數(shù)相乘。
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 Rn 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
編譯器C++ 源碼:
#include <iostream>
#include <string>
using namespace std;
#define MAX 255
int getnum(string s,int *c) // get real number of R
{
int i=0,j=0,t[MAX];
memset(t,0,sizeof(int)*MAX); // a stores 0
while (i < 6) // R value 1 through 6
{
if (s[i] != '.')
{
t[j]=s[i]-'0';
j++;
}
i++;
} // a`s length = 5
for (j=0; j<5; j++)
c[j]=t[4-j]; // c stores in order from a
for (i=0; s[i] != '.'; i++); // find decimal point
return (5-i); // the position of . point
}
void multi(int *a,int *b) // big-multiplication
{
int i=0,j,r=0,t[MAX];
memset(t,0,sizeof(int)*MAX); // t stores 0
for (; i<5; i++)
for (j=0; j<255; j++)
t[i+j] += a[i]*b[j]; // core algorithms!
for (i=0; i<255; i++)
{
b[i]=(r+t[i])%10; // r always stores remainder
r=(r+t[i])/10; // b stores the result
}
} // basic algorithms of b-m
int main()
{
int i,j,d_pos,n,a[MAX],b[MAX];
string s;
while (cin>>s>>n)
{
memset(b,0,sizeof(int)*MAX);
memset(a,0,sizeof(int)*MAX);
d_pos=getnum(s,a);
getnum(s,b);
for (i=0; i<n-1; i++)
multi(a,b); // a is a loop invariant
for (i=254; !b[i]; i--); //find last non-zero
for (j=0; !b[j]; j++); // find first non-zero
for (; i >= n*d_pos; i--) // loop n times
cout<<b[i];
if (n*d_pos >= j+1) cout<<"."; //pay attention
for (i=n*d_pos-1; i>=j; i--)
cout<<b[i]; //from back formating output
cout<<endl;
}
return 0;
}