hdu 1002（大数）

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 318065    Accepted Submission(s): 61825

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

 

字符串数组模拟

#include<iostream>
#include<cstring>
#include<string.h>
#include<cmath>
#include<algorithm>
typedef long long ll;
using namespace std;
#define N 1010
int main()
{
    int n,k,i,j,flag;
    char s1[N],s2[N];
    int a[N],b[N];
    int c[N];
    cin>>n;
    for(k=1;k<=n;k++)
    {
        cin>>s1>>s2;
        int len1=strlen(s1);
        int len2=strlen(s2);
        cout<<"Case "<<k<<":"<<endl<<s1<<" + "<<s2<<" = ";
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        int j=0;
        int jj=0;
        for(i=len1-1;i>=0;i--)
        {
            a[j]=s1[i]-'0';
            j++;
        }
        for(i=len2-1;i>=0;i--)
        {
            b[jj]=s2[i]-'0';
            jj++;
        }
        memset(c,0,sizeof(c));
        for(i=0;i<=max(len1-1,len2-1);i++)
        {
            c[i]=a[i]+b[i]+c[i];
            if(c[i]>9)
            {
                c[i]=c[i]-10;
                c[i+1]++;
            }
        }
        flag=0;
        for(i=max(len1-1,len2-1);i>=0;i--)
        {
            if(flag==0&&c[i]==0)
                flag=0;
            else
            {
                cout<<c[i];
                flag=1;
            }
        }
        if(k<n)
            cout<<endl<<endl;
        else
            cout<<endl;
    }
    return 0;
}