高精度算法总结(C++)
四种大整数运算
- 核心:将A的每一位放在数组里,且第0位存放个位数字,即逆序存放
高精度加法
- 核心:
- 参考题目:LeetCode989
#include <iostream>
#include <cstring>
#include <vector>
using namespace std;
const int N = 10e6 + 10;
// 模板
vector<int> add(vector<int>& A, vector<int>& B)
{
vector<int> C;
int t = 0; // 进位
for (int i = 0; i < A.size() || i < B.size(); i++){
if (i < A.size()) t += A[i]; // 当前为有A[i] 就加上 B[i]同理,得到当前为相加为A[i] + B[i] + t
if (i < B.size()) t += B[i];
C.push_back(t % 10);
t /= 10;
}
if (t) C.push_back(1);
return C;
}
int main()
{
string a, b;
vector<int> A, B;
cin >> a >> b;
for (int i = a.size() - 1; i >= 0; i--) A.push_back(a[i] - '0');
for (int i = b.size() - 1; i >= 0; i--) B.push_back(b[i] - '0');
auto C = add(A, B);
for (int i = C.size() - 1; i >= 0; i--) printf("%d", C[i]);
return 0;
}
高精度减法
假设两个正整数相减
- 核心: 或
#include <iostream>
#include <vector>
#include <cstring>
using namespace std;
const int N = 1e5 + 10;
// 判断 A >= B
bool cmp(vector<int>& A, vector<int>& B)
{
if (A.size() != B.size()) return A.size() > B.size();
else{
for (int i = A.size() - 1; i >= 0; i--) // 位数相同再比较每一位的大小
if (A[i] != B[i])
return A[i] > B[i];
}
return true;
}
vector<int> _sub(vector<int>& A, vector<int>& B)
{
vector<int> C;
int t = 0;
for (int i = 0; i < A.size(); i++){
t = A[i] - t;
if (i < B.size()) t -= B[i]; // 如果B存在
C.push_back((t + 10) % 10);
if (t < 0) t = 1;
else t = 0;
}
while (C.size() > 1 && C.back() == 0) C.pop_back(); // 去掉前导0
return C;
}
int main()
{
string a, b;
cin >> a >> b;
vector<int> A, B;
for (int i = a.size() - 1; i >= 0; i--) A.push_back(a[i] - '0');
for (int i = b.size() - 1; i >= 0; i--) B.push_back(b[i] - '0');
if (cmp(A, B)){
auto C = _sub(A, B);
for (int i = C.size() - 1; i >= 0; i--) printf("%d", C[i]);
}
else{
auto C = _sub(B, A);
printf("-");
for (int i = C.size() - 1; i >= 0; i--) printf("%d", C[i]);
}
return 0;
}
高精度乘法
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
const int N = 1e6 + 10;
vector<int> mul(vector<int>& A, int b)
{
vector<int> C;
int t = 0;
for (int i = 0; i < A.size() || t; i++){ // i没循环完或进位没处理完
if (i < A.size()) t += A[i] * b;
C.push_back(t % 10);
t /= 10;
}
return C;
}
int main()
{
string a;
int b;
cin >> a >> b;
vector<int> A;
for (int i = a.size() - 1; i >= 0; i--) A.push_back(a[i] - '0');
auto C = mul(A, b);
for (int i = C.size() - 1; i >= 0; i--) printf("%d", C[i]);
return 0;
}
高精度除法
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
const int N = 1e6 + 10;
// A / b,商是C,余数是r
vector<int> dev(vector<int>& A, int b, int& r)
{
vector<int> C;
r = 0;
for (int i = A.size() - 1; i >= 0; i--){
r = r * 10 + A[i];
C.push_back(r / b);
r %= b;
}
reverse(C.begin(), C.end());
while (C.size() > 1 && C.back() == 0) C.pop_back();
return C;
}
int main()
{
string a;
int b;
cin >> a >> b;
vector<int> A;
for (int i = a.size() - 1; i >= 0; i--) A.push_back(a[i] - '0');
int r;
auto C = dev(A, b, r);
for (int i = C.size() - 1; i >= 0; i--) printf("%d", C[i]);
cout << endl << r << endl;
return 0;
}
来源:CSDN
作者:孤岛violet
链接:https://blog.csdn.net/fly_wt/article/details/103653708