A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 315232 Accepted Submission(s): 61142

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

Author

Ignatius.L

题解：大数A+B，直接用模板来解决（附上不同的代码）

#include <iostream>
#include <cstring>
using namespace std;
const int M=1000+10;
string add(string a,string b)
{
    string c;
    int len1=a.length();
    int len2=b.length();
    int len=max(len1,len2);
    for(int i=len1;i<len;i++)
        a="0"+a;
    for(int i=len2;i<len;i++)
        b="0"+b;
    int ok=0;
    for(int i=len-1;i>=0;i--)
    {
        char temp=a[i]+b[i]-'0'+ok;
        if(temp>'9')
        {
            ok=1;
            temp-=10;
        }
        else ok=0;
        c=temp+c;
    }
    if(ok) c="1"+c;
    return c;
}
int main()
{
 int t;
 string a,b;
 cin>>t;
 for(int i=1;i<=t;i++)
 {
  if(i>1) cout<<endl;
  cin>>a>>b;
  cout<<"Case "<<i<<":"<<endl;
  cout<<a<<" + "<<b<<" = "<<add(a,b)<<endl;

 }
 return 0;
}

View Code

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;
const int M=100000+10;
string add(char * a,char * b)
{

    char d[M]="",e[M]="";
    int len1=strlen(a);
    int len2=strlen(b);
    int len=max(len1,len2);
    d[0]=e[0]='0';
    for(int i=len1,j=1;i<len;i++)
        d[j++]='0';
    strcat(d,a);
    for(int i=len2,j=1;i<len;i++)
        e[j++]='0';
    strcat(e,b);
    //cout<<"d="<<d<<",e="<<e<<endl;
    len=strlen(d);//cout<<"len="<<len<<endl;
     for(int i=len-1;i>=0;i--)
    {
      int x=d[i]-'0',y=e[i]-'0';
      if(x+y>9)
      {
       d[i-1]+=1;
       d[i]+=y-10;
       }
      else   {d[i]+=y;}
    }
     if (d[0]=='0')return (d+1);
     else     return d;
}
int main()
{
 int t;
 char a[M],b[M];
 cin>>t;
 for(int i=1;i<=t;i++)
 {
  if(i>1) printf("
");
  scanf("%s%s",a,b);
  cout<<"Case "<<i<<":"<<endl;
  cout<<a<<" + "<<b<<" = "<<add(a,b)<<endl;

 }
 return 0;
}

View Code

#include<cstdio>
#include<cstring>
using namespace std;
int main()
{
     int j=1,p=0,i,n,aa,bb;
     char a[1000],b[1000],c[1000];
     scanf("%d",&n);
     while(n)
     {
          scanf("%s%s",a,b);
          printf("Case %d:
",j++);
          printf("%s + %s = ",a,b);
          aa=strlen(a)-1;bb=strlen(b)-1;
          for( i=0;aa>=0||bb>=0;aa--,bb--,i++)
          {
               if(aa>=0&&bb>=0)c[i]=a[aa]+b[bb]-'0'+p;
               if(aa>=0&&bb<0)c[i]=a[aa]+p;
               if(aa<0&&bb>=0)c[i]=b[bb]+p;
               if(c[i]>'9'){c[i]=c[i]-10;p=1;}
               else p=0;
          }
          if(p==1)printf("%d",p);
          while(i--)printf("%c",c[i]);
          if(n==1) printf("
");
          else printf("

");
          n--;
     }
     return 0;
}

View Code