HDOJ1002

/*
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
*/

#include//高精度加法
#include
#define MAX 1004
using namespace std;
int main(){
    int a_length,b_length;//a,b的长度
    int len,x,t=0;//len是sum数组长度，x是进位
    int i[MAX],j[MAX];//转存成数字数组
    char a[MAX],b[MAX];//为了好输入呀
    int sum[MAX];//相加结果数组
    int testnumber;//有几组测试数据
    cin>>testnumber;
    for(int p=1;p<=testnumber;p++){
        memset(i,0,sizeof(i));
        memset(j,0,sizeof(j));
        memset(sum,0,sizeof(sum));
    //    memset(a,0,sizeof(a));
    //    memset(b,0,sizeof(b));
        if(t>0) cout<>a; cin>>b;
        a_length=strlen(a);
        b_length=strlen(b);
        
        for(int k=0;k0;jj--) cout<

其实就是高精度加法吧，然后我半天格式错误，因为题目要非第一次输入的时候在输入前面有个回车，这就是t的用处啦