【演算法】【高精度】大數相關問題總結
阿新 • • 發佈:2020-07-16
常考
大數乘法
題目連結:https://leetcode-cn.com/problems/multiply-strings/
class Solution { public: string multiply(string num1, string num2) { string res(num1.size() + num2.size(), '0'); for(int i = num1.size() - 1; i >= 0; i--) for(int j = num2.size() - 1; j >= 0; j--) { int val = (res[i + j + 1] - '0') + (num1[i] - '0') * (num2[j] - '0'); res[i + j + 1] = val % 10 + '0'; res[i + j] += val / 10; } for(int i = 0; i < res.size(); i++) if(res[i] != '0') return res.substr(i); return "0"; } };
大數加法
題目連結:https://leetcode-cn.com/problems/add-strings/
class Solution { public: string addStrings(string num1, string num2) { string res; reverse(num1.begin(), num1.end()); reverse(num2.begin(), num2.end()); int val = 0; for(int i = 0; i < num1.size() || i < num2.size(); i++) { if(i < num1.size()) val += num1[i] - '0'; if(i < num2.size()) val += num2[i] - '0'; res.push_back(val % 10 + '0'); val /= 10; } if(val) res.push_back(val + '0'); reverse(res.begin(), res.end()); return res; } };
大數減法
#include <iostream> #include <string> #include <vector> using namespace std; bool cmp(vector<int> &A, vector<int> &B) { if(A.size() != B.size()) return A.size() > B.size(); for(int i = A.size() - 1; i >= 0; i--) if(A[i] != B[i]) return A[i] > B[i]; return true; } vector<int> sub_(vector<int> &A, vector<int> &B) { vector<int> C; for(int i = 0, t = 0; i < A.size(); ++i) { t = A[i] - t; if(i < B.size()) t -= B[i]; C.push_back((t + 10) % 10); if(t < 0) t = 1; else t = 0; } while(C.size() > 1 && C.back() == 0) C.pop_back(); 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'); vector<int> C; if(cmp(A, B)) C = sub_(A, B); else C = sub_(B, A), printf("-");; for(int i = C.size() - 1; i >= 0; --i) printf("%d", C[i]); }
大數除法
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;
vector<int> div_(vector<int> &A, int b, int &r)
{
vector<int> C;
int t = 0;
for(int i = A.size() - 1; i >= 0; i--)
{
t = t * 10 + A[i];
C.push_back(t / b);
t %= b;;
}
r = t;
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 = 0;
auto C = div_(A, b, r);
for(int i = C.size() - 1; i >= 0; --i) cout<<C[i];
cout<<endl;
cout<<r<<endl;
}
n的階乘(考慮溢位)
#include <iostream>
#include <vector>
#include <string>
using namespace std;
vector<int> mul(vector<int> A, int b)
{
vector<int> C;
int t = 0;
for(int i = 0; i < A.size() || t; ++i)
{
if(i < A.size()) t += A[i] * b;
C.push_back(t % 10);
t /= 10;
}
return C;
}
int main()
{
int n;
cin >> n;
vector<int> A;
A.push_back(1);
for(int i = 1; i <= n; i++)
{
A = mul(A, i);
}
for(int i = A.size() - 1; i >= 0; i--) cout<<A[i];
cout<<endl;
}