HDU1002

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 360342    Accepted Submission(s): 69944

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

21 2112233445566778899 998877665544332211

 

Sample Output

Case 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110
//高精度加法 将字符串转换为整型数组（逆序），即两个数组相加的时候从低位向高位加
#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>using namespace std;int c[1001];int main(int argc, const char * argv[]) {    int n;    cin >> n;    int count = 1;    while(n--)    {        string a,b;        cin >> a >> b;        int m[1001] = {0},l[1001] = {0};                int num_a = 0,num_b = 0;        for(long i = a.size()-1;i >= 0; --i)            m[num_a++] = a[i] - '0';        for(long i = b.size()-1;i >= 0; --i)            l[num_b++] = b[i] - '0';        int len = max(num_a,num_b);        memset(c, 0, 1000*sizeof(int));        for(int i = 0;i <= len; ++i)//单独用数组存储，进行相加后进位处理        {            c[i] += l[i] + m[i];            if(c[i] >= 10)            {                c[i] %= 10;                c[i+1] = 1;            }        }        cout << "Case " << count++ << ":" << endl;        cout << a << " + " << b << " = ";        if(c[len] == 0)            len = len -1;        for(int i = len;i >= 0; --i)            cout << c[i];        cout << endl;        if(n > 0)            cout << endl;    }    return 0;}