hdoj 1002 A + B Problem II【大数加法】
A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 260585 Accepted Submission(s): 50389
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> #define MAX 1100 char str1[MAX],str2[MAX]; int a[MAX],b[MAX]; int main() { int n,m,j,i,s,t,l1,l2,k,ok; scanf("%d",&t); k=1; while(t--) { memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); scanf("%s%s",str1,str2); l1=strlen(str1); l2=strlen(str2); for(i=l1-1,j=0;i>=0;i--) { a[j]=str1[i]-'0'; j++; } for(i=l2-1,j=0;i>=0;i--) { b[j]=str2[i]-'0'; j++; } n=0; if(l1<l2) { n=l1; l1=l2; l2=n; } for(i=0;i<l1;i++) { a[i]=a[i]+b[i]; if(a[i]>=10) { a[i]-=10; a[i+1]++; } } ok=0; if(a[l1]>0) { ok=1; } printf("Case %d:\n",k++); printf("%s + %s = ",str1,str2); if(!ok) for(i=l1-1;i>=0;i--) printf("%d",a[i]); else for(i=l1;i>=0;i--) printf("%d",a[i]); printf("\n"); if(t) printf("\n"); } return 0; }