A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 244704    Accepted Submission(s): 47183

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

解题

#include
    
     #include
     
      int main(){    int n,m,len1,len2,maxlen,minlen;    char a[1010],b[1010],sum[1010];    char *max, *min;    scanf("%d",&n);    for(m=1;m<=n;m++)    {        int i,j,flag=0;        memset(sum,0,sizeof(sum));        scanf("%s %s",a,b);        len1=strlen(a);        len2=strlen(b);        if(len1>=len2)//判断字符串哪个长        {            maxlen=len1;            minlen=len2;            max = a;            min = b;        }        else        {            maxlen=len2;            minlen=len1;            max = b;            min = a;        }        for(i=maxlen-1,j=minlen-1;i>=maxlen-minlen&&j>=0;i--,j--)//做循环把两字符串相加放到sum,用flag记录进位,大于9时候为1，并且sum-10，        {                                                       //只加完到最短的(从后面开始加，方便输出从前面0下表开始输出)            sum[i]+=max[i]+min[j]-'0'-'0';            if(i==0&&sum[0]>9)            {                flag=1;                sum[0]-=10;            }            else if(sum[i]>9)            {                sum[i-1]+=1;//进位                sum[i]-=10;//减10            }        }        for(i=maxlen-minlen-1;i>=0;i--)//再加比最短的长出的部分的,从后面开始加。        {            sum[i]+=max[i]-'0';            if(i==0&&sum[0]>9)            {                flag=1;                sum[i]-=10;            }            else if(sum[i]>9)            {                sum[i-1]++;//进位                sum[i]-=10;//减10            }        }        // printf("       %d\n",flag);        if(flag)                    //输出格式        {            printf("Case %d:\n",m);            printf("%s + %s = 1",a,b);}//a，b字符串不能修改，flag=1，sum[0]=0的时候的进位        else        {            printf("Case %d:\n",m);            printf("%s + %s = ",a,b);        }        for(i=0;i