A+B Problem（高精）（板子？）

「爱情、让人受尽委屈。」 2022-12-13 04:22 79阅读 0赞

#### 题目描述 ####

高精度加法，相当于a+b problem，不用考虑负数.

#### 输入格式 ####

分两行输入。a,b≤10500

#### 输出格式 ####

输出只有一行，代表a+b的值

##### 输入输出样例 #####

输入

1001
    9099

输出

10100

#include<iostream>
    #include<cstring>
    using namespace std;
    int main()
    { 
        string s1,s2;
        int a[250],b[250],len;
        int i;
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        cin>>s1>>s2;
        a[0]=s1.length();
        b[0]=s2.length();
        for(i=1;i<=a[0];i++)
        { 
      	a[i]=s1[a[0]-i]-'0'; 
        } 
        for(i=1;i<=b[0];i++)
        { 
      	b[i]=s2[b[0]-i]-'0';
        }
        len=(a[0]>b[0]?a[0]:b[0]);
        for(i=1;i<=len;i++)
        { 
      	a[i]+=b[i];
      	a[i+1]+=a[i]/10;
      	a[i]%=10;
        }
        len++;
        while((a[len]==0)&&(len>1)) len--;
        for(i=len;i>=1;i--)
        { 
      	cout<<a[i];
        }
        cout<<endl;
        return 0;
    }

或

//运算符重载
    #include<iostream>
    #include<string>
    #include<cstring>
    #include<cstdio>
    using namespace std;
    const int N = 1005;
    struct bign
    { 
        int len,s[N];
        bign()  {   memset(s,0,sizeof(s));  len=1;  }
        bign(int num)  {   *this=num; }
        bign(char *num) {  *this=num; }
        bign operator =(int num)
        { 
            char c[N];
            sprintf(c,"%d",num);
            *this=c;
            return *this;
        }
        bign operator =(const char *num)
        { 
            len=strlen(num);
            for (int i=0;i<len;i++) s[i]=num[len-1-i]-'0';
            return *this;
        }
        string str()
        { 
            string res="";
            for (int i=0;i<len;i++) res=(char)(s[i]+'0')+res;
            return res;
        }
        void clean()
        { 
            while (len>1&&!s[len-1]) len--;
        }
        bign operator +(const bign &b)
        { 
            bign c;    
            c.len=0;
            for (int i=0,g=0;g||i<len||i<b.len;i++)
            { 
                int x=g;
                if (i<len) x+=s[i];
                if (i<b.len) x+=b.s[i];
                c.s[c.len++]=x%10;
                g=x/10;
            }
            return c;
        }
        bign operator -(const bign &b)
        { 
            bign c;
            c.len=0;
            int x;     
            for (int i=0,g=0;i<len;i++)
            { 
                x=s[i]-g;
                if (i<b.len) x-=b.s[i];
                if (x>=0) g=0;
                else{           
                    x+=10;
                    g=1;
                };
                c.s[c.len++]=x;
            }
            c.clean();
            return c;
        }
        bign operator *(const bign &b)
        { 
            bign c;
            c.len=len+b.len;
            for (int i=0;i<len;i++) for (int j=0;j<b.len;j++) c.s[i+j]+=s[i]*b.s[j];
            for (int i=0;i<c.len-1;i++) {  c.s[i+1]+=c.s[i]/10; c.s[i]%=10; }
            c.clean();
            return c;  
        }
        bool operator <(const bign &b)
        { 
            if (len!=b.len) return len<b.len;
            for (int i=len-1;i>=0;i--)
                 if (s[i]!=b.s[i]) return s[i]<b.s[i];
            return false;
        }
        bign operator +=(const bign &b)
        { 
            *this=*this+b;
            return *this;
        }
        bign operator -=(const bign &b)
        { 
            *this=*this-b;
            return *this;
        }  
    };
    istream &operator >>(istream &in,bign &x)
    { 
        string s;
        in>>s;
        x=s.c_str();
        return in;
    }
    ostream &operator <<(ostream &out,bign &x)
    { 
        out<<x.str();
        return out;
    }
    int main()
    { 
        bign a,b,c;
        ios::sync_with_stdio(false);
        cin>>a>>b;
        c=a+b;
        cout<<c<<endl;
        return 0;
    }

或

#include<cstdio>
    #include<cstring>
    #include<iostream>
    using namespace std;
    int compare(string s1,string s2)
    { 
        if(s1.length()>s2.length()) return 1;
        else if(s1.length()<s2.length()) return -1;
        else return s1.compare(s2);
    }
    //高精度加法
    string add(string str1,string str2)
    { 
        string str;
        int len1=str1.length();
        int len2=str2.length();
        if(len1<len2)	//前面补0，弄成长度相同
        { 
            for(int i=1;i<=len2-len1;i++)
               str1="0"+str1;
        }
        else
        { 
            for(int i=1;i<=len1-len2;i++)
               str2="0"+str2;
        }
        len1=str1.length();
        int cf=0;
        int temp;
        for(int i=len1-1;i>=0;i--)
        { 
            temp=str1[i]-'0'+str2[i]-'0'+cf;
            cf=temp/10;
            temp%=10;
            str=char(temp+'0')+str;
        }
        if(cf!=0)  str=char(cf+'0')+str;
        return str;
    }
    int main()
    { 
        string a,b;
        cin>>a>>b;
        cout<<add(a,b)<<endl;
    }

或

#include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<cstring>
    const int wei = 4, maxn = 1e6+10, mod = 10000;
    using namespace std;
    char t1[maxn], t2[maxn];
    int a[maxn], b[maxn], ans[maxn];
    int main()
    { 
        scanf("%s%s", t1+1, t2+1);
        int len1 = strlen(t1+1), len2 = strlen(t2+1);
        int _len1 = len1 / wei, _len2 = len2 / wei;
        if(len1 % wei) _len1++; 
        if(len2 % wei) _len2++;
        int w;
        for(int i=1; i<=_len1; i++)
        { 
            w = 1;
            for(int j=1; j<=wei && (i-1)*wei+j <= len1; j++)
      	{ 
                a[i] += w * (t1[len1-(i-1)*wei-j+1] - '0');
                w *= 10;
            }
        }
        for(int i=1; i<=_len2; i++)
        { 
            w = 1;
            for(int j=1; j<=wei && (i-1)*wei+j <= len2; j++)
      	{ 
                b[i] += w * (t2[len2-(i-1)*wei-j+1] - '0');
                w *= 10;
            }
        }
        int _len = max(_len1, _len2);
        for(int i=1; i<=_len; i++)
        { 
            ans[i] += a[i] + b[i];
            ans[i+1] += ans[i] / mod;
            ans[i] %= mod;
        }
        int top = _len+1; 
        if(ans[top] == 0) top--;
        cout << ans[top];
        for(int i=top-1; i>0; i--) printf("%0*d", wei, ans[i]);
        return 0;
    }

或

//结构体版本
    #include<bits/stdc++.h>
    using namespace std;
    int x,y;
    int i,j;
    char ab[10001],bc[100001];
    int a[10001],b[100001];
    int jw=0;
    int maxn,maxj;
    inline void read()
    { 
        cin>>ab>>bc;
        x=strlen(ab);
        y=strlen(bc);
        for(i=0;i<x;i++) a[i]=ab[x-i-1]-'0';
        for(i=0;i<y;i++) b[i]=bc[y-i-1]-'0';
    }
    inline void solve()
    { 
        maxn=max(x,y);
        for(i=0;i<maxn;i++)
        { 
      	a[i]=a[i]+b[i]+jw;
      	jw=a[i]/10;
      	a[i]=a[i]%10;
        }
        if(jw==1) cout<<jw;
        for(i=maxn-1;i>=0;i--) cout<<a[i];
    }
    int main()
    { 
        read();
        solve();
        return 0;
    }