uva748 - Exponentiation 高精度小数的幂运算
| 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 .
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
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
套用模版，注意小数的位数不足时要补0，后续0要清掉（其实对底数进行后续0清除就好了，因为出现后续0的唯一可能就是底数有后续0）
/* 高精度的幂。幂为低精度。 */ #include <cstdio> #include <iostream> #include <cstring> #include <climits> using namespace std; #define maxn 30000 struct bign { int len, s[maxn]; bign() { memset(s, 0, sizeof(s)); len = 1; } bign(int num) { *this = num; } bign(const char* num) { *this = num; } bign operator = (int num) { char s[maxn]; sprintf(s, "%d", num); *this = s; return *this; } bign operator = (const char* num) { len = strlen(num); for (int i = 0; i < len; i++) s[i] = num[len-i-1] - '0'; return *this; } string str() const { string res = ""; for (int i = 0; i < len; i++) res = (char)(s[i] + '0') + res; if (res == "") res = "0"; return res; } bign operator + (const bign& b) const { bign c; c.len = 0; for (int i = 0, g = 0; g || i < max(len, b.len); i++) { int x = g; if (i < len) x += s[i]; if (i < b.len) x += b.s[i]; c.s[c.len++] = x % 10; g = x / 10; } return c; } void clean() { while(len > 1 && !s[len-1]) len--; } bign operator * (const bign& b) { bign c; c.len = len + b.len; for (int i = 0; i < len; i++) for (int j = 0; j < b.len; j++) c.s[i+j] += s[i] * b.s[j]; for (int i = 0; i < c.len-1; i++) { c.s[i+1] += c.s[i] / 10; c.s[i] %= 10; } c.clean(); return c; } bign operator - (const bign& b) { bign c; c.len = 0; for (int i = 0, g = 0; i < len; i++) { int x = s[i] - g; if (i < b.len) x -= b.s[i]; if (x >= 0) g = 0; else { g = 1; x += 10; } c.s[c.len++] = x; } c.clean(); return c; } bool operator < (const bign& b) const { if (len != b.len) return len < b.len; for (int i = len-1; i >= 0; i--) if (s[i] != b.s[i]) return s[i] < b.s[i]; return false; } bool operator > (const bign& b) const { return b < *this; } bool operator <= (const bign& b) { return !(b > *this); } bool operator == (const bign& b) { return !(b < *this) && !(*this < b); } bool operator != (const bign& b) { return (b < *this) || (*this < b); } bign operator += (const bign& b) { *this = *this + b; return *this; } }; istream& operator >> (istream &in, bign& x) { string s; in >> s; x = s.c_str(); return in; } ostream& operator << (ostream &out, const bign& x) { out << x.str(); return out; } int main() { bign a,ans; int b; string c; while (cin >> c >> b) { ans = 1; int index = c.find('.'); if (index != -1) { //后续0 for (int i = c.size()-1; i > index; --i) { if (c[i] == '0') c.erase(i, 1); else break; } c.erase(index, 1); } index = c.size()-index; index *= b; a = c.c_str(); a.clean(); for (int i = 0; i < b; ++i) { ans = ans*a; } string s = ans.str(); //补0 int dif = s.size()-index; if (dif >= 0) { s.insert(s.end()-index, '.'); } else { for (int i = 0; i < -dif; ++i) { s = '0' + s; } s = '.' + s; } cout << s << endl; } }