大数相加(hdu1022)

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 348933    Accepted Submission(s): 67756

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

#include<stdio.h>#include<string.h>int main(){    int t,j=1,flag2=0;    scanf("%d",&t);    int a[10010],b[10010],c[10010];    char str1[10010],str2[10010];    while(t--)    {        if(flag2==1)            printf("\n");        memset(a,0,sizeof(a));        memset(b,0,sizeof(b));        memset(c,0,sizeof(c));        scanf("%s%s",str1,str2);        int k,i;        int len1,len2,len_max;        len1=strlen(str1);        len2=strlen(str2);        for(i=0; i<len1; i++)            a[i]=str1[len1-1-i]-'0';        for(i=0; i<len2; i++)            b[i]=str2[len2-1-i]-'0';        if(len1>len2)            len_max=len1;        else            len_max=len2;        k=0;        for(i=0; i<len_max; i++)        {            c[i]=(a[i]+b[i]+k)%10;            k=(a[i]+b[i]+k)/10;        }        if(k!=0)            c[len_max]=1;        printf("Case %d:\n",j);        j++;        printf("%s + %s = ",str1,str2);        if(c[len_max]==1)            printf("%d",c[len_max]);        for(i=len_max-1; i>=0; i--)            printf("%d",c[i]);        printf("\n");        flag2=1;    }    return 0;}