hdu 1002 解析

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 317305    Accepted Submission(s): 61642

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

题意：输入T组测试数据，没组测试数据输入两个数。

注意：输入的数据很大。所以不能直接用int来解决。

思路：用数组来处理。代码看起来长，思路其实很简单。

代码如下：

#include<iostream>#include<string.h>#include<cstdio>using namespace std;int main(){    int n,nn=1;    cin>>n;    while(n!=0)    {        char aa[10000],bb[10000];        int a[10000],b[10000],c[10000];        memset(a,0,sizeof(a));//对a数组清零        memset(b,0,sizeof(b));        memset(c,0,sizeof(c));        cin>>aa>>bb;//用字符数组才能把数字一位一位的存储        int la=strlen(aa),lb=strlen(bb);        for(int i=0;i<la;i++)            a[i]=aa[la-i-1]-'0';//颠倒了位置，个位数在a【0】的位置了        for(int i=0;i<lb;i++)            b[i]=bb[lb-i-1]-'0';        int l;        l=max(la,lb);        int bl=0;        for(int i=0;i<l;i++)        {            int bll=a[i]+b[i]+bl;//bl上次相加是进的位            if(bll>=10)//要是相加大于10，就进位            {                bl=1;                bll=bll-10;            }            else bl=0;            c[i]=bll;        }        if(bl==1)//判断最大位的相加是否会大于10        {            c[l]=1;            cout<<"Case "<<nn<<":"<<endl;            cout<<aa<<" + "<<bb<<" = ";            for(int i=l;i>=0;i--)            cout<<c[i];            cout<<endl;        }        else        {            cout<<"Case "<<nn<<":"<<endl;            cout<<aa<<" + "<<bb<<" = ";            for(int i=l-1;i>=0;i--)            cout<<c[i];             cout<<endl;        }        n--;nn++;        if(n!=0)            cout<<endl;//每组数据之间要有空行，最后一组不要有！    }    return 0;}