HDU 1002 A + B Problem II

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 367839    Accepted Submission(s): 71655

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

 

思路：大数相加问题。将两个数组中的字符-‘0’，模拟加法运算，判断其相加的结果是否大于10，结果大于10则使当前位减10、下一位加1。要注意细节的处理，如输出格式以及多种情况的判别。

#include <iostream>#include<stdio.h>#include<string.h>using namespace std;int main(){    int n,i,len1,len2,j,k,pi,t,flag;    char str1[1010],str2[1010];    cin>>n;    for(i=1;i<=n;i++)    {        int a[1200]={0};        flag=0;        printf("Case %d:\n",i);        scanf("%s%s",str1,str2);//以字符串形式读入        len1=strlen(str1);        len2=strlen(str2);        printf("%s + %s = ",str1,str2);        j=len1-1;        k=len2-1;        pi=0;        while(j>=0&&k>=0)//开始相加        {            if(a[pi]+(str1[j]-'0')+(str2[k]-'0')>=10)//相加后大于等于10的            {                a[pi]+=(str1[j]-'0')+(str2[k]-'0')-10;                a[pi+1]++;            }            else//相加后小于10的                a[pi]+=(str1[j]-'0')+(str2[k]-'0');            pi++;            k--;            j--;        }        if(j>=0)//str1字符数较大时        {            for(t=j;t>=0;t--)            {                a[pi]+=(str1[t]-'0');                pi++;            }        }        else if(k>=0)//tr2字符数较大时        {            for(t=k;t>=0;t--)            {                a[pi]+=str2[t]-'0';                pi++;            }        }        else if(a[pi]!=0)//对于字符数相同的2个数加后最高位大于10的            pi++;        for(t=pi-1;t>=0;t--)        {            if(a[t]==0&&flag==0)//处理前导0，flag标记是否处理完成                continue;            else            {                flag=1;                printf("%d",a[t]);            }        }        printf("\n");        if(i!=n)//处理2组结果之间加空行的情况            printf("\n");    }    return 0;}