HDU 1002.A + B Problem II-数组模拟-大数相加

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 435932    Accepted Submission(s): 84825

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

 

 

Author

Ignatius.L

 

 

 

代码:

#include<stdio.h>
#include<string.h>
int main(){
    char a[1005],b[1005];
    int c[1005],d[1005],ans[1005];
    int n,i,k,j,h,t,l;
    int len1,len2;
    scanf("%d",&n);
        h=0;
        while(n--){
            h+=1;
            for(i=0;i<1005;i++){
                c[i]=d[i]=0;
                ans[i]=0;
            }
            scanf("%s %s",&a,&b);
            len1=strlen(a);
            len2=strlen(b);
            for(i=0;i<len1;i++){
               c[i]=a[i]-'0';
            }
            for(i=0;i<len2;i++){
               d[i]=b[i]-'0';
            }
            printf("Case %d:\n",h);
            for(i=0;i<len1;i++)
                printf("%d",c[i]);
                printf(" + ");
            for(i=0;i<len2;i++)
                printf("%d",d[i]);
                printf(" = ");
            for(i=0,j=len1-1;i<len1/2;i++,j--){
                t=c[j];
                c[j]=c[i];
                c[i]=t;
            }
            for(i=0,j=len2-1;i<len2/2;i++,j--){
                t=d[j];
                d[j]=d[i];
                d[i]=t;
            }
            l=len1>len2?len1:len2;
            for(i=0,k=0;i<=l;i++,k++){
                if(c[i]+d[i]<10)
                    ans[k]=c[i]+d[i];
                else{
                    ans[k]=(c[i]+d[i])%10;
                    c[i+1]+=(c[i]+d[i])/10;
                }

            }
            for(j=k-1;j>=0;j--){
                if(j==k-1&&ans[j]==0)
                    continue;
                printf("%d",ans[j]);
            }
                if(n==0)
                    printf("\n");
                else
                    printf("\n\n");
        }
    return 0;
}