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A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9348    Accepted Submission(s): 1557


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
长整数加法,程序如下:
#include<iostream>
using namespace std;

#define Base 10
#define MaxLen  1000

struct BigInt {
    
int len;
    
int data[MaxLen];
    
    BigInt() :len(
0{}
    BigInt(
const BigInt& s) :len(s.len){
        memcpy(
this->data,s.data,len*sizeof(*data));
    }

    BigInt(
int s) :len(0){
        
for(;s>0;s=s/Base)
            data[len
++]=s%Base;
    }


    BigInt 
& operator = (const BigInt& s){
        
this->len=s.len;
        memcpy(
this->data,s.data,len*sizeof(*data));
        
return *this;
    }


    
int& operator [](int index)return data[index]; }
    
int operator [](int index) const return data[index]; }
}
;

BigInt 
operator + (BigInt & a, BigInt & b){
    BigInt c;
    
int i,carry=0;
    c.len
=a.len>b.len?a.len:b.len;
    
for(i=0;i<c.len||carry>0;++i){
        
if(i<a.len)    carry+=a[i];
        
if(i<b.len)    carry+=b[i];
        c[i]
=carry%Base;
        carry
/=Base;
    }

    c.len
=i;
    
return c;
}


istream 
& operator >> (istream & is, BigInt & s)
{
    
int i;
    
char ch[MaxLen];
    
is>>ch;
    s.len
=strlen(ch);
    
for(i=s.len-1;i>-1;--i)
        s[i]
=ch[s.len-1-i]-'0';
    
return is;
}


ostream 
& operator << (ostream & os, BigInt & s)
{
    
int i;
    
for(i=s.len-1;i>-1;--i)
        os
<<s[i];
    
return os;
}



int main () {
    
int n,i;
    BigInt a,b,c;
    
    cin
>>n;
    
for(i=1;i<n;++i){
        cin
>>a>>b;
        c
=a+b;
        cout
<<"Case "<<i<<":"<<endl;
        cout
<<a<<" + "<<b<<" = "<<c<<endl<<endl;
    }

    cin
>>a>>b;
    c
=a+b;
    cout
<<"Case "<<i<<":"<<endl;
    cout
<<a<<" + "<<b<<" = "<<c<<endl;
    
return 0;
}


此处需注意输出的问题,即最后一组A+B不用输出2个换行符(仔细看题),否则PE。
posted on 2007-11-28 18:57 tctony 阅读(702) 评论(1)  编辑 收藏 引用

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# re: HDU 1002 2011-05-17 21:12 bell
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