hdoj 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): 293743    Accepted Submission(s): 56546

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

 

Author

Ignatius.L

 

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大数相加，不过刚开始在细节上出了个小错误，找了好长时间

#include<stdio.h>#include<string.h>int main(){int n,i,j,k,l,t,p,c[1000],d[1000];char a[1100],b[1100];scanf("%d",&n);t=1;while(n--){scanf("%s",a);scanf("%s",b);printf("Case %d:\n",t++);printf("%s + %s = ",a,b);memset(c,0,sizeof(c));memset(d,0,sizeof(d));k=strlen(a);l=strlen(b);j=0;for(i=k-1;i>=0;i--)c[j++]=a[i]-'0';p=0;for(i=l-1;i>=0;i--)d[p++]=b[i]-'0';for(i=0;i<1100;i++){c[i]=c[i]+d[i];if(c[i]>=10){c[i]=c[i]-10;c[i+1]++;}}for(i=999;i>=0;i--)if(c[i]!=0)break;for(j=i;j>=0;j--)printf("%d",c[j]);printf("\n");if(n!=0)printf("\n");}return 0; }