A + B Problem II---hdu1002 A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 260647 Accepted Submission(s): 50397
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
1 #include<stdio.h> 2 #include<string.h> 3 #define as 1000 4 int main() 5 { 6 int as1[as+20],as2[as+20],cot=0,j,t; 7 char shuru1[as+20],shuru2[as+20]; 8 int n,i; 9 scanf("%d",&n); 10 getchar();//吸收回车符 11 t=n; 12 while(n--) 13 { 14 cot++; 15 scanf("%s",shuru1); 16 scanf("%s",shuru2); 17 memset(as1,0,sizeof(as1)); 18 memset(as2,0,sizeof(as2)); 19 for(i=0,j=strlen(shuru1)-1;j>=0;j--,i++) 20 { 21 as1[i]=shuru1[j]-'0'; 22 23 } 24 for(i=0,j=strlen(shuru2)-1;j>=0;j--,i++) 25 { 26 as2[i]=shuru2[j]-'0'; 27 } 28 for(i=0;i<as+20;i++) 29 { 30 as1[i]+=as2[i]; 31 if(as1[i]>=10)//判断是否满十进一 32 { 33 as1[i+1]++; 34 as1[i]-=10; 35 } 36 37 } 38 for(i=as+19;(i>=0)&&(as1[i]==0);i--);//去掉结果前面多余的0 39 printf("Case %d: ",cot); 40 printf("%s + %s = ",shuru1,shuru2); 41 if(i>=0) 42 { 43 for(;i>=0;i--) 44 printf("%d",as1[i]); 45 } 46 else 47 printf("0"); 48 printf(" "); 49 if(cot!=t) 50 printf(" ");//最后不要多换行 51 } 52 return 0; 53 }