HDU 1002 A + B Problem II
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A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 329420 Accepted Submission(s): 63949
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
21 2112233445566778899 998877665544332211
Sample Output
Case 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110
Author
Ignatius.L
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大数加法
ac代码:
#include <stdio.h>#include <string.h>int main(){char a[1005],b[1005];int t,len_a,len_b,i,min,jinwei;int x,y,tmp,cnt=0;scanf("%d",&t);getchar();while(t--){scanf("%s%s",a,b);getchar();len_a=strlen(a);len_b=strlen(b);printf("Case %d:\n",++cnt);printf("%s + %s = ",a,b);jinwei=0;min=len_a<len_b?len_a:len_b;if(len_a>=len_b){for(i=min-1;i>=0;i--){x=a[--len_a]-'0';y=b[i]-'0';tmp=(x+y+jinwei)%10;a[len_a]=tmp+48;jinwei=(x+y+jinwei)/10;}if(len_a>0)while(jinwei){x=a[--len_a]-'0';tmp=(x+jinwei)%10;a[len_a]=tmp+48;jinwei=(x+jinwei)/10;}else if(jinwei)printf("%d",jinwei);puts(a);}else{for(i=min-1;i>=0;i--){x=a[i]-'0';y=b[--len_b]-'0';tmp=(x+y+jinwei)%10;b[len_b]=tmp+48;jinwei=(x+y+jinwei)/10;}if(len_b>0)while(jinwei){x=b[--len_b]-'0';tmp=(x+jinwei)%10;b[len_b]=tmp+48;jinwei=(x+jinwei)/10;}else if(jinwei)printf("%d",jinwei);puts(b);}if(t>0)printf("\n");}return 0;}
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