标签:mem out turn += class const mat 过程 printf
虽然noip不会重点考这个但是为了防止不必要的丢分还是准备一下比较好。
高精度的原理就是我们小学的时候所使用的竖式运算。
所以以字符串的形式模拟一下运算过程结果就出来了。
#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<string> #include<cmath> #include<algorithm> using namespace std; const int MAXN = 410; 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 = (const int num) { char t[MAXN]; sprintf(t, "%d", num); *this = t; return *this; } bign operator = (const char *num) { for(int i = 0; num[i] == ‘0‘; num++); len = strlen(num); for(int i = 0; i < len; i++) s[i] = num[len - i - 1] - ‘0‘; return *this; } bign operator + (const bign &b) { 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; } bign operator += (const bign &b) { *this = *this + b; return *this; } 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; i++) { c.s[i + 1] = c.s[i] / 10; c.s[i] %= 10; } c.clean(); return c; } bign operator *= (const bign &b) { *this = *this * b; return *this; } 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; } bign operator -= (const bign &b) { *this = *this - b; return *this; } bign operator / (const bign &b) { bign c, f = 0; for(int i = len - 1; i >= 0; i--) { f = f * 10; f.s[0] = s[i]; while(f >= b) { f -= b; c.s[i]++; } } c.len = len; c.clean(); return c; } bign operator /= (const bign &b) { *this = *this / b; return *this; } bign operator % (const bign &b) { bign r = *this / b; r = *this - r * b; return r; } bign operator %= (const bign &b) { *this = *this % b; return *this; } bool operator < (const bign &b) { 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) { 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) { return !(*this > b) && !(*this < b); } bool operator >= (const bign &b) { return *this > b || *this == b; } bool operator <= (const bign &b) { return *this < b || *this == b; } bool operator != (const bign &b) { return *this > b || *this < b; } string str() const { string ret = ""; for(int i = len - 1; i >= 0; i--) ret += char(s[i] + ‘0‘); return ret; } }; istream& operator >> (istream &in, bign &x) { string s; in >> s; x = s.c_str(); return in; } ostream& operator << (ostream &out, bign &x) { out << x.str(); return out; } int main() { bign a, b, c, d;//可加 cin >> a >> b >> d;//可改 c = a - b + d;//可改 cout << c; return 0; }
以上为封装好的高精度。
请各位大爷尽情享用。
一世安宁
标签:mem out turn += class const mat 过程 printf
原文地址:https://www.cnblogs.com/GTBA/p/9451497.html