A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 287467 Accepted Submission(s): 55255
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
2
1 2
112233445566778899 998877665544332211
Sample Output
Case 1:
1 + 2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110
#include<stdio.h> #include<string.h> int main() { int n,j; scanf("%d",&n); for(j=0;j<n;j++) { int a[1200]={0},b[1200]={0},c[1200]={0},i,long1,long2,k; char a1[1200],b1[1200]; scanf("%s",a1); scanf("%s",b1); long1=strlen(a1); long2=strlen(b1); for(i=0;i<long1;i++) a[i]=a1[long1-i-1]-'0'; for(i=0;i<long2;i++) b[i]=b1[long2-i-1]-'0'; k=long1<long2?long2:long1; for(i=0;i<k;i++) c[i]=a[i]+b[i]; for(i=0;i<k;i++) if(c[i]>=10) { c[i+1]+=c[i]/10; c[i]%=10; } printf("Case %d: ",j+1); printf("%s + %s = ",a1,b1); if(c[k]==0) k--; for(i=k;i>=0;i--) printf("%d",c[i]); if(j==n-1) printf(" "); else printf(" "); } return 0; }