hdu 1003 A + B Problem II 使用整型數組輕鬆實現大數求和

爲題如下:

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
 
 
 
解體代碼如下,實現很簡單,這裏就不說明實現過程
 
 
  1. /********************************************** 
  2.     > File Name: 1002.c 
  3.     > Author: sea 
  4.     > Mail: [email protected] 
  5.     > Created Time: Mon 06 May 2013 02:01:19 PM CST 
  6.  **********************************************/ 
  7.  
  8. #include<stdio.h> 
  9. #include<string.h> 
  10.  
  11. #define N 1101 
  12.  
  13. int main() { 
  14.     char str1[N],str2[N]; 
  15.     int sum[N+1],i,j,m,n,t,s1,s2,k,flag=1; 
  16.     scanf("%d",&i); 
  17.     while(i--) { 
  18.         k=N-1; 
  19.         memset(str1,0,sizeof(str1)); 
  20.         memset(str2,0,sizeof(str2)); 
  21.         memset(sum,0,sizeof(sum)); 
  22.         scanf("%s %s",str1,str2); 
  23.         m=strlen(str1)-1; 
  24.         n=strlen(str2)-1; 
  25.         if(m>n) t=m; 
  26.         else t=n; 
  27.         for(j=0;j<=t;j++,m--,n--,k--) { 
  28.             if(m<0) s1=0; 
  29.             else s1=str1[m]-48; 
  30.             if(n<0) s2=0; 
  31.             else s2=str2[n]-48; 
  32.             if(sum[k]+s1+s2>=10) 
  33.                 sum[k-1]+=1; 
  34.             sum[k]=(sum[k]+s1+s2)%10; 
  35.  
  36.         } 
  37.         sum[k]>0?k:k++; 
  38.         printf("Case %d:\n%s + %s = ",flag++,str1,str2); 
  39.         for(j=k;j<N;j++) 
  40.             printf("%d",sum[j]); 
  41.         if(i>=1)    printf("\n\n"); 
  42.         else printf("\n"); 
  43.     } 
  44.     return 0; 
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