POJ-北大acm 1001 exponentiation 解题报告与bug

     严格来说本解法不能算大数相乘的通用算法, 而只是针对题目中(0.0 < R < 99.999)的trick. 因R位数很小, 所以本算法思路是把R扩大倍数而转成整数(99.999 -> 99999), 然后把每次乘积累加到结果数组中, 算法比较简单.

     同时, 发现POJ本题的审核程序有bug. 题目要求对尾部(trailing)无效的0不能打印出来, 但实际审核时没有这么严谨, 可能是测试数据没有涵盖所有边界或者其它原因. 具体情况是: 对小于0的结果打印时不去掉尾部的0, 审核程序依然accepted. 举个例子, 对于小于1的R(如0.10000), 不论n取多少(如取3), 其乘方的结果是小于0的小数`.001000000000000`, 如果算法直接打印`.001000000000000`依然accepted, 但依题目的要求真正结果应是`.001`. 即把后文算法中如下注释掉的4行代码打开才是严格的, 但POJ的bug导致注释了也能accepted.

//while (product[j] == 0) // trim trailing zeros
//{
//	j++;
//}

以下是题目与解法(C语言), 文末贴了参考来源

                                          Exponentiation

Time Limit: 500MS		Memory Limit: 10000K
Total Submissions: 116340		Accepted: 28271

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>

int len; // total length of exponentiation result
int product[126] = {0}; // storing result, at most length 5*25 + 1 = 126

void multiply(int a[], int n)
{
    int i;
    int carry = 0; // a carry number in multiplying
    for (i = 0; i < len; i++)
    {
        int temp = a[i]*n + carry;
        a[i] = temp % 10;
        carry = temp / 10;      
    }
    while (carry)
    {
        a[i++] = carry % 10;
        carry /= 10;
    }
    len = i;
}

int main(int argc, char* argv[])
{
    int n;  // power n
    char s[6]; // real number R, at most the length is 6
    while (scanf("%s %d", s, &n) != EOF)
    {
        int position=0, i=0, num=0, j=0;
        for (i=0; i<strlen(s); i++) 
        {
            if (s[i] == '.')
            {
                position = (strlen(s) - 1 - i) * n; // calculate decimal point position after R^n
            }
            else
            {
                num = num*10 + s[i] - 48; // transfer float to integer
            }       
        }
        
        // product calculation 
        product[0]=1;
        len = 1;
        for (i = 0; i < n; i++)
        {
            multiply(product, num);
        }

        // format output
        if (len <= position) // product is less than 1
        {
            printf("."); // print decimal point
            for (i=0; i<position-len; i++)
            {
                printf("0"); // print zero between decimal point and decimal
            }

            j = 0;
            //while (product[j] == 0) // trim trailing zeros
            //{
            //    j++;
            //}
            for (i=len-1; i>=j; i--)
            {
                printf("%d", product[i]);
            }
        }   
        else
        {
            j=0;
            while (product[j]==0 && j<position) // trim trailing zeros
            {
                j++;
            }
            for (i=len-1; i>=j; i--)
            {
                if (i+1 == position) // cause index in C language starts from 0
                {
                    printf(".");
                }
                printf("%d", product[i]);
            }
        }
        printf("\n");
    }
}

参考: http://blog.csdn.net/xiongheqiang/article/details/7471661