HDOJ 题目1002A + B Problem II （大数）

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 209712    Accepted Submission(s): 40343

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

 题目大意：大数的加法

ac代码

#include<stdio.h>#include<string.h>int main(){int cot=0,n;char a[1010],b[1010];scanf("%d",&n);while(n--){int c[1100],len1,len2,i,j,k=0,max;getchar();scanf("%s %s",a,b);len1=strlen(a);len2=strlen(b);i=len1-1;j=len2-1;memset(c,0,sizeof(c));while(i>=0||j>=0){if(i<0&&j>=0)c[k]+=b[j]-'0';elseif(i>=0&&j<0)c[k]+=a[i]-'0';elsec[k]+=(a[i]-'0')+(b[j]-'0');k++;c[k]=c[k-1]/10;c[k-1]%=10;if(c[k])max=k;elsemax=k-1;j--;i--;}printf("Case %d:\n",++cot);printf("%s + %s = ",a,b);for(j=max;j>=0;j--)printf("%d",c[j]);printf("\n");if(n)printf("\n");}}