杭电oj-1002-A+B Problem

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. <br />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 <br />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

问题分析

此题是有一定难度的，其难点在于，如何表示当有1000位数的数字，因为就算是long long 型也是无法表达这么大的数字的（long long 是能表示出最大的数字的数据类型，其范围为9223372036854775807 ~ -9223372036854775808），因此应考虑采用数组进行存储，把我们输入的数转换成字符串，再将其转换为整数，另外，在本题中要注意相加的几种情况，

第一种，A.length==B.length

第二种，A.length>B.length

第三种，A.length<B.length

对于每种情况应该如何考虑，最后要注意题中所给的输入、输出格式。

Code：

#include<iostream>
#include<string>
#include<cstdio>
using namespace std;
string a,b;
int sum[1005],q;//q为相加之后的位数 
void add2(string a,string b) {
        /*两个大数相加求和*/
    int m,n,t=0,i=0,int_a,int_b;
    //t为进位数 
    m=a.length()-1;
    n=b.length()-1;
    while(m>=0&&n>=0){
        int_a=a[m]-'0';//将输入的每一位数先转换成int型 
        int_b=b[n]-'0';
        sum[i]=(int_a+int_b+t)%10;//相加后第i位的数字
        t=(int_a+int_b+t)/10;//相加后向上一位进的位数
        i++; 
        m--; 
        n--; 
    }
    if(m>n){
        while(m>=0){
            int_a=a[m]-'0';
            sum[i]=(int_a+t)%10;
            t=(int_a+t)/10;
            m--;
            i++;
        }
        
    }
    if(m<n){
        while(n>=0){
            int_b=b[n]-'0';
            sum[i]=(int_b+t)%10;
            t=(int_b+t)/10;
            n--;
            i++;    
        }
            
    }
    q=i-1;
    if(t>=1)//最后一次相加存在进位时 
        {
            sum[i]=t;
            q=i;
        }
    

}
void output() {
    /*输出函数*/
    int i=q;
    cout<<a<<" + "<<b<<" = ";
    for(;i>=0;i--){
        
        cout<<sum[i];        
    }
    cout<<endl; 
    
}
int main() {
    int n;//测试用例数
    cin>>n;
    int i=0;
    for(; i<n; i++) {
        cin>>a>>b;
        add2(a,b);
        printf("Case %d:\n",i+1);//测试样例格式 
        output();
        if(i<n-1)
        cout<<endl;
    }
    return 0;
}