A + B Problem II

HDU 1002 ：*A + B Problem II*

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

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

本题题意：
  本题意思是给你两个数要你计算两数之和。不过这题是大数加法问题。

解题思路：
  这是大数的加法问题，用字符数组来存储大数，然后转化为整形数组来计算即可。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <sstream>#define max(a,b) a>b?a:busing namespace std;const int Max=1000;void getDigits(int num[],char str[])///把字符串保存在int类型的数组中{    char digit;    int len=strlen(str);    for(int i=0; i<len; ++i)    {        digit=str[i];        num[len-i-1]=digit-'0';///字符串倒序保存    }}void add(int a[],int b[],int sum[]){    for(int i=0; i<Max; i++)    {        sum[i]=a[i]+b[i];    }    ///将十位以上的数字向上进位，将剩余的数字保存在自己的位置上    for(int i=0; i<Max; i++)    {        sum[i+1]+=sum[i]/10;        sum[i]=sum[i]%10;    }}int main(){    char num1[Max],num2[Max];    int a[Max];    int b[Max];    int sum[Max*2];    int n,m;    cin>>n;    for(m=1; m<=n; m++)    {        scanf("%s%s",&num1,&num2);        getchar();        memset(a,0,sizeof(a));        memset(b,0,sizeof(b));        memset(sum,0,sizeof(sum));        getDigits(a,num1);        getDigits(b,num2);        add(a,b,sum);        int j=Max*2-1;0        while(sum[j]==0)        {            j--;        }        cout<<"Case "<<m<<":"<<endl;        cout<<num1<<" + "<<num2<<" = ";        for(int i=j; i>=0; i--)        {            printf("%d",sum[i]);        }        printf("\n");        if(m!=n)            printf("\n");    }    return 0;}