c++高精度模板

古城微笑少年丶 2022-08-29 10:51 146阅读 0赞

两个大整数相乘

#include <iostream>
    #include <cstring>
    using namespace std;
    
    const int N = 50000;
    char a[N], b[N];
    int A[N], B[N], C[2*N];
    
    int main()
    { 
        cin >> a >> b;
        
        A[0] = strlen(a);
        B[0] = strlen(b);
        
        for(int i=1; i <= A[0]; i++) A[i] = a[A[0]-i] - '0';
        
        for(int j=1; j <= B[0]; j++) B[j] = b[B[0]-j] - '0';
    
        for(int i = 1; i <= A[0]; i++)
            for(int j = 1; j <= B[0]; j++)
                C[i+j-1] += A[i] * B[j];
                
        int len = A[0] + B[0];
        
        for(int i = 1; i < len; i++)
        { 
            if(C[i] > 9) 
            { 
                C[i+1] += C[i] / 10;
                C[i] %= 10;
            }
        }
        
       while(C[len] == 0 && len > 1) len--;
        
        for(int i = len; i >= 1; i--)
            printf("%d", C[i]);
            
        return 0;
    }

## 高精度加法 ##

两个大整数做加法

//高精度加法
    #include <iostream>
    #include <vector>
    
    using namespace std;
    
    vector<int> add(vector<int> &A, vector<int> &B)
    { 
        vector<int> C;
        int t = 0;
        for(int i = 0; i<A.size()||i<B.size(); i++)
        { 
            if(i < A.size()) t += A[i];
            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;
    
        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');
    
        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 ##

两个大整数做减法

/********* sub **********/
    #include <iostream>
    #include <bits/stdc++.h>
    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;
    }

--------------------

## 高精度除法 ##

大数除小数，取商

#include <iostream>
    #include <algorithm>
    #include <vector>
    using namespace std;
    
    vector<int> div(vector<int> &a, int &b)
    { 
        vector<int> c;
    
        long 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());
    
        return c;
    }
    
    int main()
    { 
        string a1;
    
        int b;
    
        cin >> a1 >> b;
    
        vector<int> a;
    
        for(int i = a1.size()-1; i >= 0; i--) 
         a.push_back(a1[i]-'0');
    
        auto c = div(a, b);
    
        while(c.back() == 0 && c.size() > 1) c.pop_back();
        for(int i = c.size()-1; i >= 0; i--)
            printf("%d",c[i]);
        
        return 0;
    }

--------------------