东华OJ-基础题-106-大整数相加（C++）

问题描述
I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.
输入说明
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.
输出说明
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.
输入范例

cpp 复制代码

2
1 2
112233445566778899 998877665544332211

输出范例

cpp 复制代码

Case 1:
1 + 2 = 3

Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110

感想：定义一个包含A与B的结构体，定义一个使用该结构体的vector,对vector里的A与B相加进行计算即可；注意输出的+左右两边得有空格，= 左右两边也得有空格，要不然PE。
代码如下：

cpp 复制代码

#include <bits/stdc++.h>
using namespace std;

struct add {
    string a;
    string b;
};

int main() {
    int n;
    cin>>n;
    vector<add> arr(n);
    for(int i = 0; i<n; ++i) {
        cin>>arr[i].a>>arr[i].b;
    }

    for(int i =0; i<n; ++i) {
        string temp_a = arr[i].a,temp_b = arr[i].b;
        string ans;
        int j = temp_a.size();
        int k = temp_b.size();
        int carry = 0,add;
        while(j&&k) {
            add = (temp_a[--j]-'0')+(temp_b[--k]-'0')+carry;;
            carry = add/10;
            ans = to_string(add%10)+ans;
        }
        while(j) {
            add = (temp_a[--j] -'0' )+ carry;
            carry = add/10;
            ans = to_string(add%10)+ans;
        }
        while(k) {
            add = (temp_a[--k] -'0' )+ carry;
            carry = add/10;
            ans = to_string(add%10)+ans;
        }

        if(carry) ans = to_string(carry) + ans;
        if(i>0) cout<<endl;
        cout<<"Case "<<i+1<<":"<<endl;
        cout<<arr[i].a<<" + "<<arr[i].b<<" = "<<ans<<endl;

    }

    return 0;
}