//思路:1.调用函数change() 将小数 s 转化成去掉小数点的整数 s1 并且返回得到小数点的位置。
// 2.调用函数mu1()用大数乘法模拟整数运算,并将相乘的结果放在s2中,返回s2和是进行下面的 n -1次相乘
// 3.在主函数里面通过已知的小数点的位置,利用数值关系输出
// 4.注意小数点前没有前导0,小数点后面没有尾0。
//难点:for ( int i = 1; i < n; i ++) mu1 (s1,s2);
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#include <stdio.h>
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#include <stdlib.h>
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#include <string.h>
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#define LENGTH 6 //小数的位数(含小数点)
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//将字符转化为数字
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unsigned int change (char s[LENGTH], unsigned int s1[LENGTH - 1])
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{
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int ss[LENGTH - 1]; //ss 放未逆置的整数
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memset (ss, 0, sizeof(ss));
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int k = 0 ;
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for (int i = 0; i < LENGTH && s[i]; ++i) //考虑特殊数据如:0.0001
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{
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if (s[i] != '.')
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ss[k++] = s[i] - '0';
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}
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for (int j = 0;j < LENGTH - 1; j++)
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{
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s1[j] = ss[LENGTH - 2 - j];
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}
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int m = 0;
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while ( (s[m] != '.') && s[m] )
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++m;
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return LENGTH - 1 - m; //小数点位数
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}
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//大数乘法运算
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//函数返回 s2
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void mu1 (unsigned int s1[LENGTH - 1],unsigned int s2[130])
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{
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int ss[130];
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memset ( ss, 0, sizeof(ss) );
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for ( int i = 0; i < LENGTH - 1; i++)
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for (int j = 0;j < 130; j++) //难点:因为返回新的s2之后位数会增加 最多时 5* 25 = 125
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ss [i + j] += s1[i] * s2[j];
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//将 两个大数相乘得的积ss中进行进位处理后放到s2 中
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int c = 0;
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for (int i = 0;i < 130;i++)
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{
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s2[i] = (c + ss[i]) % 10;
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c = (c + ss[i]) / 10;
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}
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}
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int main()
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{
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int n;
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char s[LENGTH]; //要处理的幂 R
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unsigned int s1[LENGTH - 1]; //将 R 转化成数字
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unsigned int s2[130];
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while(scanf ("%s%d", s, &n) != EOF)
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{
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memset (s1, 0, sizeof (s1));
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memset (s2, 0, sizeof (s2));
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int j = change (s, s1); //得到小数点所在位置
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change (s,s2); //得到s2 和 s1 进行幂运算
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for ( int i = 1; i < n; i ++)
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mu1 (s1,s2);
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//在s2中前面的代表小数位,后面的代表整数位,
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//所以关键是通过数值关系找到小数点的位置
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//例:0.1010 * 0.1010 = 0.01020100
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int m = 129;//去掉前导0
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while ( (!s2[m]) && m)
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--m;
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int k = 0; //去掉尾0
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while ( ( !s2[k] ) && (k < 130))
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++k;
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//输出整数位
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for (int i = m; i >= n * j; i--)
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printf ("%d",s2[i]);
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//输出小数点
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if ( j && n * j >= k + 1)
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printf (".");
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for (int i = n*j -1; i >= k; --i)
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printf ("%d", s2[i]);
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printf ("\n");
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}
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return 0;
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// system ("pause");
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}
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posted on 2010-08-09 13:21
雪黛依梦 阅读(615)
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