HDU 1002 - A + B Problem II - [高精度]

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1002

Problem Description
I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.

Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000.

Output
For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line is the an equation "A + B = Sum", Sum means the result of A + B. Note there are some spaces int the equation. Output a blank line between two test cases.

Sample Input
2
1 2
112233445566778899 998877665544332211

Sample Output
Case 1:
1 + 2 = 3

Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110

 

题意：

给出 $A,B$，求 $A+B$。

 

题解：

数据范围太大，需要手工模拟加法。

 

AC代码：

#include<stdio.h>
#include<string.h>
#define max(a,b) ((a>b)?a:b)
int main()
{
    char ta[5001],tb[5001];
    int a[5001],b[5001],sum[5002],n,t,i,lena,lenb,lenmax;
    scanf("%d",&n);
    for(t=1;t<=n;t++)
    {
        scanf("%s %s",ta,tb);
        lena=strlen(ta);
        lenb=strlen(tb);
        lenmax=max(lena,lenb);
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b)); 
        memset(sum,0,sizeof(sum));
        for(i=0;i<=lena-1;i++) a[i]=ta[lena-1-i]-'0';
        for(i=0;i<=lenb-1;i++) b[i]=tb[lenb-1-i]-'0';
        for(i=0;i<=lenmax-1;i++)
        {
            sum[i]+=(a[i]+b[i]);
            if(sum[i]>=10)
            {
                if(i==lenmax-1) lenmax++;
                sum[i]-=10;
                sum[i+1]++;
            }
        }
        printf("Case %d:\n",t);
        printf("%s + %s = ",ta,tb);
        for(i=lenmax-1;i>=0;i--) printf("%d",sum[i]); printf("\n");
        if(t<n) printf("\n");
    }
}

（默默把当年刚入坑时敲得C语言代码折叠）拿出我们的高精度板子！

#include<bits/stdc++.h>
using namespace std;
const int maxn=1005;

struct BigInt
{
    int len,d[maxn];

    void clean(){while(len>1 && !d[len-1]) len--;}
    void output(){for(int i=len-1;i>=0;i--) printf("%d",d[i]);}
    string str()const
    {
        string s;
        for(int i=0;i<len;i++) s+=d[len-1-i]+'0';
        return s;
    }

    BigInt(){memset(d,0,sizeof(d));len=1;}
    BigInt(int num){*this=num;}
    BigInt(char* num){*this=num;}

    bool operator<(const BigInt& oth)const
    {
        if(len!=oth.len) return len<oth.len;
        for(int i=len-1;i>=0;i--) if(d[i]!=oth.d[i]) return d[i]<oth.d[i];
        return false;
    }
    bool operator>(const BigInt& oth)const{return oth<*this;}
    bool operator<=(const BigInt& oth)const{return !(oth<*this);}
    bool operator>=(const BigInt& oth)const{return !(*this<oth);}
    bool operator!=(const BigInt& oth)const{return oth<*this || *this<oth;}
    bool operator==(const BigInt& oth)const{return !(oth<*this) && !(*this<oth);}

    BigInt operator=(const char* num)
    {
        memset(d,0,sizeof(d));
        len=strlen(num);
        for(int i=0;i<len;i++) d[i]=num[len-1-i]-'0';
        clean();
        return *this;
    }
    BigInt operator=(int num)
    {
        char s[20];
        sprintf(s,"%d",num);
        return *this=s;
    }
    BigInt operator+(const BigInt& oth)const
    {
        BigInt c;
        c.len=max(len,oth.len);
        for(int i=0;i<=c.len;i++) c.d[i]=0;
        for(int i=0;i<c.len;i++)
        {
            c.d[i]+=(i<len?d[i]:0)+(i<oth.len?oth.d[i]:0);
            c.d[i+1]+=c.d[i]/10;
            c.d[i]%=10;
        }
        c.len+=(c.d[c.len]>0);
        c.clean();
        return c;
    }
    BigInt operator-(const BigInt& oth)const
    {
        BigInt c=*this;
        if(c<oth) printf("Produce negative number!\n");
        int i;
        for(i=0;i<oth.len;i++)
        {
            c.d[i]-=oth.d[i];
            if(c.d[i]<0) c.d[i]+=10, c.d[i+1]--;
        }
        while(c.d[i]<0) c.d[i++]+=10, c.d[i]--;
        c.clean();
        return c;
    }
    BigInt operator*(const BigInt& oth)const
    {
        BigInt c;
        for(int i=0;i<len;i++) for(int j=0;j<oth.len;j++) c.d[i+j]+=d[i]*oth.d[j];
        for(int i=0;i<len+oth.len || !c.d[i];c.len=++i) c.d[i+1]+=c.d[i]/10, c.d[i]%=10;
        c.clean();
        return c;
    }
    BigInt operator/(const BigInt& oth)const
    {
        BigInt c=*this, r=0;
        for(int i=0;i<len;i++)
        {
            r=r*10+c.d[len-1-i];
            int j;
            for(j=0;j<10;j++) if(r<oth*(j+1)) break;
            c.d[len-1-i]=j;
            r=r-oth*j;
        }
        c.clean();
        return c;
    }
    BigInt operator%(const BigInt& oth)
    {
        BigInt r=0;
        for(int i=0;i<len;i++)
        {
            r=r*10+d[len-1-i];
            int j;
            for(j=0;j<10;j++) if(r<oth*(j+1)) break;
            r=r-oth*j;
        }
        return r;
    }
    BigInt operator+=(const BigInt& oth)
    {
        *this=*this+oth;
        return *this;
    }
    BigInt operator*=(const BigInt& oth)
    {
        *this=*this*oth;
        return *this;
    }
    BigInt operator-=(const BigInt& oth)
    {
        *this=*this-oth;
        return *this;
    }
    BigInt operator/=(const BigInt& oth)
    {
        *this=*this/oth;
        return *this;
    }
}a,b;

char A[maxn],B[maxn];
int main()
{
    int T;
    cin>>T;
    for(int kase=1;kase<=T;kase++)
    {
        scanf("%s%s",A,B);
        a=A, b=B;
        printf("Case %d:\n",kase);
        a.output();
        printf(" + ");
        b.output();
        printf(" = ");
        (a+b).output();
        printf("%s",kase<T?"\n\n":"\n");
    }
}