杭电 1002 大数相加 【关键语句：sum=a[i]-'0'+b[j]-'0'+carry;c[i]='0'+sum%10;//-‘0’字符串转换成数字-‘0’，数字转换成字符串+‘0’】

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 324547    Accepted Submission(s): 63050

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

#include
#include
int main()
{
    char a[1024],b[1024],c[1024];
    int n,t,l1,l2,l,i,j,carry,sum;
    while(EOF!=scanf("%d",&n))
    {
        for(t=0;t<n;t++)
        {
            if(t!=0)
                printf("\n");
            carry=0;
            memset(a,0,sizeof(a));
            memset(b,0,sizeof(b));
            memset(c,0,sizeof(c));
            scanf("%s%s",&a,&b);
            
            l1=strlen(a);l2=strlen(b);
            
            if(l1>l2)
            {    
                for(i=l1-1,j=l2-1;j>=0;i--,j--)
                {
                    sum=a[i]-'0'+b[j]-'0'+carry;//由字符串转换为数字——-‘0”
                    //这句还有个error，就是自动默认了i和j相等= =
                    c[i]='0'+sum%10;//由数字转换为字符串
                    carry=sum/10;
                }
                for(;i>=0;i--)//处理剩下一部分
                {
                    sum=a[i]-'0'+carry;
                    c[i]='0'+sum%10;
                    carry=sum/10;
                }
            }
            else
            {
                for(i=l1-1,j=l2-1;i>=0;i--,j--)
                {
                    sum=a[i]-'0'+b[j]-'0'+carry;
                    c[j]='0'+sum%10;
                    carry=sum/10;
                }
                for(;j>=0;j--)
                {
                    sum=b[j]-'0'+carry;
                    c[j]='0'+sum%10;//和上面的不太一样啊...
                    carry=sum/10;
                }
            }
            l=l1>l2?l1:l2;
            printf("Case %d:\n",t+1);
            printf("%s + %s = ",a,b);
            if(carry!=0)  printf("%d",carry);
            printf("%s\n",c);
            
        }
    }
    return 0;
}

#include
#include
int main()
{
    char a[1024],b[1024],c[1024];
    int n,t,l1,l2,l,i,j,carry,sum;
    while(EOF!=scanf("%d",&n))
    {
        for(t=0;tl2)
            {    
                for(i=l1-1,j=l2-1;j>=0;i--,j--)
                {
                    sum=a[i]-'0'+b[j]-'0'+carry;//由字符串转换为数字——-‘0”
                    //这句还有个error，就是自动默认了i和j相等= =
                    c[i]='0'+sum%10;//由数字转换为字符串
                    carry=sum/10;
                }
                for(;i>=0;i--)//处理剩下一部分
                {
                    sum=a[i]-'0'+carry;
                    c[i]='0'+sum%10;
                    carry=sum/10;
                }
            }
            else
            {
                for(i=l1-1,j=l2-1;i>=0;i--,j--)
                {
                    sum=a[i]-'0'+b[j]-'0'+carry;
                    c[j]='0'+sum%10;
                    carry=sum/10;
                }
                for(;j>=0;j--)
                {
                    sum=b[j]-'0'+carry;
                    c[j]='0'+sum%10;//和上面的不太一样啊...
                    carry=sum/10;
                }
            }
            l=l1>l2?l1:l2;
            printf("Case %d:\n",t+1);
            printf("%s + %s = ",a,b);
            if(carry!=0)  printf("%d",carry);
            printf("%s\n",c);
            
        }
    }
    return 0;
}