hdoj 1002 A + B Problem II 【大数加法模板】

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 265911    Accepted Submission(s): 51497

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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思路：

 

          大数加法模板题！

 

代码：

 

#include <stdio.h>#include <string.h>char a[1005];char b[1005];int a1[1005];int b1[1005];int main(){int T;int flag=0;scanf("%d",&T);int N=T;while(T--){memset(a1,0,sizeof(a1));//一定要清零 memset(b1,0,sizeof(b1));//一定要清零 if(flag==0)//注意格式！ flag=1;elseprintf("\n");scanf("%s%s",a,b);printf("Case %d:\n",N-T);printf("%s + %s = ",a,b);int lena=strlen(a);int lenb=strlen(b);int t=0;for(int i=lena-1;i>=0;i--){a1[t++]=a[i]-'0';}t=0;for(int i=lenb-1;i>=0;i--){b1[t++]=b[i]-'0';}for(int i=0;i<1005;i++){a1[i]+=b1[i];if(a1[i]>=10){a1[i+1]+=a1[i]/10;a1[i]%=10;}}int i;for(i=1005;i>=0&&a1[i]==0;i--);if(i>=0){for(;i>=0;i--){printf("%d",a1[i]);}}elseprintf("0");printf("\n");}return 0;}