字符串 高精度计算

 

原因：

　　因为不喜欢紫书的大数，所以自己写了个能跑的。

 

大数相加

#include <iostream>
#include <vector>
#include <cstring>
#include <string>
using namespace std;
int main() {
    vector<int> c;
    string s1,s2;
    cin >> s1 >> s2;
    int re = 0;
    int s = 0;
    int p1,p2;
    p1 = s1.length()-1;
    p2 = s2.length()-1;
    while(p1 >=0 || p2>=0 ) {
        s = re;
        if(p1 >= 0)    s+= s1[p1--] - '0';
        if(p2 >= 0)    s+= s2[p2--] - '0';
        re = s/10;
        c.push_back( s%10 );
    }
    if(re)    c.push_back(re);
    for(int i=c.size() - 1; i>0 && c[i]==0; --i) {
        c.pop_back();
    }
    for(int i=c.size()-1;i>=0;--i){
        cout << c[i];
    }
    return 0;
}

 

大数相减

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <vector>
using namespace std;
int main(){
    string s1,s2;
    vector<int> c;
    cin >> s1 >> s2;
    bool neg = false;
    if( (s1.length() < s2.length()) || ( (s1.length() == s2.length())&&(s1 < s2) ) ){
        neg = true;
    }
    string &a = (!neg?s1:s2),&b = (!neg?s2:s1);
    int lena = a.length()-1;
    int lenb = b.length()-1;
    int re = 0;
    int s = 0;
    while(lena >=0 || lenb >=0){
        s = re;
        re = 0;
        if(lena >= 0)    s+= a[lena--] - '0';
        if(lenb >= 0)    s-= b[lenb--] - '0';
        if(s < 0){
            re = -1;
            s += 10;            
        }
        c.push_back(s);
    }
    for(int i=c.size()-1;i>0 && c[i]==0;--i){
        c.pop_back();
    }        
    if(neg) cout << "-";
    for(int i=c.size()-1;i>=0;--i){
        cout << c[i];
    }
    return 0;
}

 

大数相乘

#include <iostream>
#include <vector>
#include <cstring>
#include <string>
using namespace std;
int main() {
    vector<int> c;
    string s1,s2;
    cin >> s1 >> s2;
    int s = 0;
    int re = 0;
    int lena = s1.length()-1;
    int lenb = s2.length()-1;
    for(int i=lena;i>=0;--i) {
        re = 0;
        for(int j=lenb;j>=0;--j){
            int ii = lena - i;//0~lena
            int jj = lenb - j;//0~lenb
        /*    [ dis ][  j  ][  i  ]
            [      c.size()        ]    */ 
            int len = c.size();
            int dis = len-jj-ii;
            s = re;
            s += (s1[i] - '0') * (s2[j] - '0');
            if(dis<=0){
                c.push_back(s%10);
            }else{
                s += c[len-dis];
                c[len-dis] = s%10;
            }
            re = s/10;
        }
        if(re){
            c.push_back(re);
        }
    }
    for(int i=c.size() - 1; i>0 && c[i]==0; --i) {
        c.pop_back();
    }
    for(int i=c.size()-1;i>=0;--i){
        cout << c[i];
    }
    return 0;
}

大数相除

　　高精度除以低精度

#include <iostream>
#include <vector>
#include <cstring>
#include <string>
#include <queue>
using namespace std;
int main() {
    string s1;
    deque<int> c;
    cin >> s1;
    int b;
    cin >> b;
    int x = 0;
    for(int i=0;i<s1.length();++i){
        int t = s1[i] - '0' + x*10;
        c.push_back( t/b );
        x = t%b;
    }
    while(c.size() >=1 && c.front() ==0){
        c.pop_front();
    }
    for(int i=0;i<c.size();i++){
        cout << c[i];
    }
    cout << endl;
    cout << x;
    return 0;
}

　　高精度除以高精度

#include <iostream>
#include <string>
#include <vector>
using namespace std;

//示范 111111/31 (可以继续优化) 
//111111
//310000
//18111
//31000
//2611
//3100
//131
//310


bool cmp(const string &s1,const string &s2) {//如果大数 a >= b 返回 true 
    return (s1.length() > s2.length()) || (s1.length()==s2.length() && s1>=s2);
}

void minus_(string &s1,const string &s2) {//大数相减 
    string ans;
    int len1 = s1.length()-1;
    int len2 = s2.length()-1;
    int s = 0;
    int re = 0;
    while(len1 >=0 || len2 >=0) {
        s = re;
        re = 0;
        if(len1 >=0)    s+=s1[len1]-'0';
        if(len2 >=0)    s-=s2[len2--]-'0';
        if(s<0) {
            re = -1;
            s+=10;
        }
        s1[len1--] = s+'0';
    }
    while( (*s1.begin()) == '0' && s1.length() > 1){
        s1.erase(s1.begin());
    }
}

void div(string s1,string s2) {
    vector<int> ans;
    int len1 = s1.length();
    int len2 = s2.length();
    int lend = len1 - len2;
    for(int i=lend;i>0;--i){
        s2.push_back('0');
    }
    int timer = 0;
    while(lend>=0){
        timer = 0;
        while(cmp(s1,s2)){
            minus_(s1,s2);
            timer++;
        }
        s2.erase(s2.end() - 1);//C11 支持pop_back();
        ans.push_back(timer);
        lend--;
    }
    for(unsigned int i=0,j=0;i<ans.size();++i){
        if(ans[i] > 0){
            j = 1;
        }
        if(j){
            cout << ans[i];
        }
    }
    cout << endl;
    cout << s1 << endl; 
}

int main() {
    string a,b;
    cin >> a >> b;
    div(a,b);
    return 0;
}