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高精度博客 - 前两讲

讲次名称	链接
高精加法	[C++] 高精度加法(作用 + 模板 + 例题)
高精乘法	[C++] 高精度乘法

高精度减法

void subBIG(int x[], int y[], int z[]){z[0] = max(x[0], y[0]);for(int i = 1; i <= z[0]; i++) z[i] = x[i] - y[i];for(int i = 1; i <= z[0]; i++){if(z[i] < 0){z[i] += 10;z[i + 1]--;}}while(z[z[0]] == 0 && z[0] > 1) z[0]--;
}

这个函数的思路还是在模拟列竖式, 最后的while是删除前缀0(以免出现类似000000的结果, 仅返回0)

高精度除法

记忆版

void divBIG(int x[],int y,int z[]){z[0] = x[0];int r = 0;for(int i = z[0]; i >= 1; i--){int t = r * 10 + x[i];z[i] = t / y;r = t % y;}while(z[z[0]] == 0 && z[0] > 1) z[0]--;
}

讲解版

void divBIG(int x[],int y,int z[]){z[0] = x[0];							// 有效位数设置// 模拟列竖式int r = 0;for(int i = z[0]; i >= 1; i--){			// 倒序循环int t = r * 10 + x[i];z[i] = t / y;r = t % y;}while(z[z[0]] == 0 && z[0] > 1) z[0]--;	// 去除前导零
}

高精度系列函数完整版

我们的3讲高精度即将结束, 给出完整的函数代码~

void s2BIG(string s, int a[]){int len = s.size();for(int i = 1; i <= len; i++){a[i] = s[len - i] - '0';}a[0] = len;
}void i2BIG(int n, int a[]){int cur = 0;while(n > 0){cur++;a[cur] = n % 10;n /= 10;}if(cur == 0) cur++;a[0] = cur;
}void printBIG(int a[]){int len = a[0];for(int i = len; i > 0; i--){cout << a[i];}cout << endl;
}void addBIG(int x[], int y[], int z[]){z[0] = max(x[0], y[0]);for(int i = 1; i <= z[0]; i++)z[i] = x[i] + y[i];for(int i = 1; i <= z[0]; i++){z[i + 1] += z[i] / 10;z[i] %= 10;if(z[z[0] + 1] != 0)z[0]++;}
}/*
如果x比y小, 则返回true
否则(y >= x)返回false
*/
bool cmpBIG(int x[], int y[]){int lx = x[0], ly = y[0];if(lx != ly) return lx < ly;for(int i = lx; i >= 1; i--){if(x[i] != y[i]) return x[i] < y[i];}return false;
}void mulBIG(int x[], int y[], int z[])
{for (int i = 1; i <= x[0]; i++){for (int j = 1; j <= y[0]; j++){z[i + j - 1] += x[i] * y[j];}}z[0] = max(x[0], y[0]);for (int i = 1; i <= z[0]; i++){z[i + 1] += z[i] / 10;z[i] %= 10;if (z[z[0] + 1] > 0) z[0]++;}while (z[z[0]] == 0 && z[0] > 1) z[0]--;
}void subBIG(int x[], int y[], int z[]){z[0] = max(x[0], y[0]);for(int i = 1; i <= z[0]; i++) z[i] = x[i] - y[i];for(int i = 1; i <= z[0]; i++){if(z[i] < 0){z[i] += 10;z[i + 1]--;}}while(z[z[0]] == 0 && z[0] > 1) z[0]--;
}void divBIG(int x[],int y,int z[]){z[0] = x[0];int r = 0;for(int i = z[0]; i >= 1; i--){int t = r * 10 + x[i];z[i] = t / y;r = t % y;}while(z[z[0]] == 0 && z[0] > 1) z[0]--;
}