A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 371689 Accepted Submission(s):
72437
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
数字太大了 只能使用字符串来保存数字
模拟我们小学学习加法的方法
从后面一位一位的加 大于9就进位
1 #include<stdio.h> 2 #include<string.h> 3 int main() 4 { 5 char op1[1002],op2[1002]; 6 int c,i,j,n,m,s1[1002],s2[1002],len1,len2,count; 7 count=1; 8 scanf("%d",&n); 9 m=n; 10 while(m--) 11 { 12 memset(s1,0,1002*sizeof(int)); 13 memset(s2,0,1002*sizeof(int)); 14 scanf("%s",op1); 15 scanf("%s",op2); 16 len1=strlen(op1); 17 len2=strlen(op2); 18 c=0; 19 for(i=len1-1;i>=0;i--) 20 s1[c++]=op1[i]-'0'; 21 c=0; 22 for(i=len2-1;i>=0;i--) 23 s2[c++]=op2[i]-'0'; 24 for(i=0;i<1002;i++) 25 { 26 s1[i]+=s2[i]; 27 if(s1[i]>=10) 28 { 29 s1[i]-=10; 30 s1[i+1]++; 31 } 32 } 33 printf("Case %d: ",count++); 34 printf("%s + %s = ",op1,op2); 35 for(i=1001;i>=0;i--) 36 if(s1[i]) 37 break; 38 for(j=i;j>=0;j--) 39 printf("%d",s1[j]); 40 printf(" "); 41 if(count!=n+1) 42 printf(" "); 43 } 44 return 0; 45 }