编程之好2.6——精确表达浮点数

2012-07-02

编程之美2.6——精确表达浮点数问题：用分数形式来表示小数，以达到精确的计算结果。解答一：用于小整数，将无限

编程之美2.6——精确表达浮点数

问题：

用分数形式来表示小数，以达到精确的计算结果。

解答一：

用于小整数，将无限循环小数0.a1a2...an(b1b2...bm)分为非循环部分和循环部分。

X=((a1a2...an)*(10^m-1)+b1b2...bm)/((10^m-1)*10^n)

#include <iostream>using namespace std;long long gcd(long long a, long long b){if (a < b){long long tmp = a;a = b;b = tmp;}int i, k=0;while (b!=0){if ((a&1) == 0){if ((b&1) == 0){// a,b均是偶数，f(a,b)=2*f(a>>1,b>>1)a >>= 1;b >>= 1;k++;}else// a为偶数，b为奇数，f(a,b)=f(a>>1,b)a >>= 1;}else{if ((b&1) == 0)// a为奇数，b为偶数，f(a,b)=f(a,b>>1)b >>= 1;else// a,b均是奇数，f(a,b)=f(a-b,b)a = a-b;}if (a < b){long long tmp = a;a = b;b = tmp;}}return a << k;}int main(){long long a=0, b=0, c=0;// 整数部分c，非循环小数a，循环小数bscanf("%d.%d(%d)",&c, &a, &b);if (a==0 && b==0)cout << c;else{// 分子up，分母downlong long up = c;long long down = 1;long long ta = a;while (ta){down *= 10;ta /= 10;}up = c*down+a;if (b!=0){long long wb = 1;long long tb = b;while (tb){wb *= 10;tb /= 10;}up = up*(wb-1)+b;down = down*(wb-1);}long long fac = gcd(up, down);cout << up/fac << "/" << down/fac << endl;}}

解答二：

用于大整数，定义了大整数类型，以及对应的加减乘除、比较移位运算。

#include <iostream>#include <cstring>#include <string>using namespace std;// 大整数类型#define MAXLEN 1000struct HP {int len, s[MAXLEN];};void PrintHP(HP x) {for (int i=x.len; i>=1; i--)cout << x.s[i];}// 字符串转大整数void Str2HP(const char *s, HP &x){x.len = strlen(s);for (int i=1; i<=x.len; i++)x.s[i] = s[x.len-i] - '0';if (x.len == 0){x.len = 1;x.s[1] = 0;}}// 大整数的加法void Plus(const HP a, const HP b, HP &c){int i; c.s[1] = 0;// 大整数a,b的加法操作和结果c的进位操作for (i=1; i<=a.len || i<=b.len || c.s[i]; i++){if (i <= a.len) c.s[i] += a.s[i];if (i <= b.len) c.s[i] += b.s[i];c.s[i+1] = c.s[i]/10; c.s[i] %= 10;}// 退出循环到原因是c.s[i]==0，所以取前一位c.len = i-1; if (c.len == 0) c.len = 1;}// 大整数的减法void Subtract(const HP a, const HP b, HP &c){int i, j;for (i=1,j=0; i<=a.len; i++){// j表示是否要对高位进行借位c.s[i] = a.s[i] - j;if (i <= b.len) c.s[i] -= b.s[i];if (c.s[i] < 0) {// 向高位借位，补10j = 1;c.s[i] += 10;}else j = 0;}c.len = a.len;while (c.len > 1 && !c.s[c.len]) c.len--;}// 大整数的比较int HPCompare(const HP &x, const HP &y){if (x.len > y.len) return 1;if (x.len < y.len) return -1;int i = x.len;while (i>1 && (x.s[i]==y.s[i])) i--;return x.s[i] - y.s[i];}// 大整数的乘法void Multi(const HP a, const HP b, HP &c){int i, j;// 对乘法结果赋初值，以方便之后的+=运算c.len = a.len + b.len;for (i=1; i<=c.len; i++) c.s[i] = 0;for (i=1; i<=a.len; i++)for (j=1; j<=b.len; j++)c.s[i+j-1] += a.s[i]*b.s[j];// 运算结果进位for (i=1; i<c.len; i++) {c.s[i+1] += c.s[i]/10; c.s[i] %= 10;}// 最高位继续进位while (c.s[i]) {c.s[i+1] = c.s[i]/10; c.s[i] %= 10; i++;}// 确保最高位不为0while (i>1 && !c.s[i]) i--;c.len = i;}// 大整数的除法void Divide(const HP a, const HP b, HP &c, HP &d){int i, j;// 用余数d存被除数a的前i位数据，用来多次减去除数b，以得到商cd.len = 1; d.s[1] = 0;for (i=a.len; i>0; i--){if (!(d.len == 1 && d.s[1] == 0)){// i没移一位，余数d也移位for (j=d.len; j>0; j--)d.s[j+1] = d.s[j];d.len++;}d.s[1] = a.s[i];c.s[i] = 0;// 余数d大于除数b时，才可以进行减操作while ((j=HPCompare(d,b)) >= 0){Subtract(d, b, d);c.s[i]++;if (j == 0) break;}}c.len = a.len;while (c.len > 1 && c.s[c.len] == 0)c.len--;}// 十进位右移void RightShift(HP &x, int k){for (int i=1; i<=x.len-k; i++)x.s[i] = x.s[i+k];x.len -= k;if(x.len <= 0){x.len = 1;x.s[1] = 0;}}// 十进位左移void LeftShift(HP &x, int k){int i;for (i=x.len; i>=1; i--)x.s[i+k] = x.s[i];for (i=k; i>=1; i--)x.s[i] = 0;x.len += k;}// 求大整数的最大公约数void GCD(HP a, HP b, HP &c){if (b.len == 1 && b.s[1] == 0){c.len = a.len;memcpy(c.s, a.s, (a.len+1)*sizeof(int));}else{HP m, n;Divide(a, b, m, n);GCD(b, n, c);}}int main(){string str;string strc, stra, strb;cin >> str;int posc = str.find('.');int posa = str.find('(');int posb = str.find(')');strc = str.substr(0, posc);if (posc < 0)cout << strc;else{HP a, b, c;HP tmp; tmp.len = 1; tmp.s[1] = 1;// 整数部分Str2HP(strc.c_str(), c);stra = str.substr(posc+1, posa-posc-1);// 非循环部分Str2HP(stra.c_str(), a);// up分子，down分母HP up = c, down = tmp;// 乘以10^|a|LeftShift(down, stra.size());LeftShift(up, stra.size());Plus(up, a, up);if (posa >= 0){strb = str.substr(posa+1, posb-posa-1);// 循环部分Str2HP(strb.c_str(), b);HP m = tmp;LeftShift(m, strb.size());Subtract(m, tmp, m);// 乘以10^(|b|-1)Multi(up, m, up);Plus(up, b, up);Multi(down, m, down);}// 求分子分母的最大公约数GCD(down, up, tmp);HP h;Divide(down, tmp, down, h);Divide(up, tmp, up, h);PrintHP(up); cout << "/";PrintHP(down); cout << endl;}}

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