大数乘法 748 - Exponentiation

最新推荐文章于 2017-09-06 15:31:46 发布

原创最新推荐文章于 2017-09-06 15:31:46 发布 · 939 阅读

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本文介绍了一个解决UVaOJ平台上的大数乘法问题的程序设计方法。该方法使用字符串来处理非常大的数值，并通过递归方式实现了幂运算。文章提供了完整的C++代码示例。

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UVa OJ

Exponentiation

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 $0 < n \le 25$ .

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 Rⁿ. Leading zeros and insignificant trailing zeros should be suppressed in the output.

Sample Input

Sample Output

548815620517731830194541.899025343415715973535967221869852721
.00000005148554641076956121994511276767154838481760200726351203835429763013462401
43992025569.928573701266488041146654993318703707511666295476720493953024
29448126.764121021618164430206909037173276672
90429072743629540498.107596019456651774561044010001
1.126825030131969720661201

这道题就是写个函数，不断进行大数乘法。用字符串来出来比较方便

还有就是我没有在每次乘完之后处理前置0。以后注意

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<ctype.h>
#include<cmath>
int digit,t,len,mul[300];
char num[10],str[10],numb[300];
using namespace std;
int exp(int t)
{
    memset(mul,0,sizeof(mul));
    int i,j,lenb=strlen(numb);
    if (t==1) return 0;
    for (i=0; i<len; i++)
    {
        for (j=0; j<=lenb; j++)
        {
            if (isdigit(num[i]) && isdigit(numb[j])) mul[i+j]+=(num[i]-'0')*(numb[j]-'0');
            if (mul[i+j]>=10)
            {
                mul[i+j+1]+=mul[i+j]/10;
                mul[i+j]%=10;
            }
        }
    }
    for (i=0; i<=len+lenb+3; i++)
        numb[i]=mul[i]+48;
    numb[i]='\0';
    exp(t-1);
}
int main ()
{
    int i,j,k;
    while(scanf("%s %d",str,&t)!=EOF)
    {
        memset(mul,0,sizeof(mul));
        memset(num,0,sizeof(num));
        memset(numb,0,sizeof(numb));
        len=strlen(str);
        for (i=len-1,j=0; i>=0; i--)
            if (isdigit(str[i])) num[j++]=str[i];
            else k=i;
        num[j]='\0';
        digit=5-k;
        len--;
        strcpy(numb,num);
        exp(t);
        digit*=t;
        for (i=strlen(numb); i>=digit; i--)
            if (numb[i]!='0' && isdigit(numb[i])) break;
        for (j=i; j>=digit; j--) printf("%c",numb[j]);
        cout<<".";
        for (i=0; ; i++)
            if (numb[i]!='0') break;
        for (j=digit-1; j>=i; j--) printf("%c",numb[j]);
        cout<<endl;

    }
    return 0;
}