A + B Problem II

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 281356 Accepted Submission(s): 54169

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<stdio.h>#include<string.h>#define N 1005int main(){    char A[N],B[N],S[N];    int T, i, j, k, l, flag;     while(scanf("%d",&T)!=EOF){ // 输入要输入的组数        for(i=0;i<T;i++){             if(i){ // 除了第一个输出，其它的都先空一行                 printf("\n");            }            scanf("%s%s",&A,&B); // A、B是两个要输入的数对应的字符串             k=strlen(A)-1; // k、l分别是A、B数组最后一位的下标             l=strlen(B)-1;            j=0,flag=0; // j是和的下标，flag是要进的数（例如：A、B某一个位和为23，则向前进2）            while((k+1)&&(l+1)){ // 取位数最小的作为求和部分（如：11+222，取11+22）                 S[j]=(A[k]-'0')+(B[l]-'0')+flag; // 求A、B之和每一位的数值                if(S[j]>=10){ // 保证每一位的合理性，如（某一位是7+2，然后他的下一位进2，那么它要向j+1位进1）                     S[j]=S[j]-10;                    flag=1;                }                else{                    flag=0;                }                 k--;                l--;                j++;            }            if(k+1){ // 非共有部分剩A数                 while(k>=0){                    S[j]=(A[k]-'0')+flag; // 第一次的flag是之前共有部分最前一位的进位                     if(S[j]>=10){ // 确定是否由于共有部分的进位引起全部进位                         S[j]=S[j]-10;                        flag=1;                    }                    else{                        flag=0;                    }                     k--;                    j++;                }            }             else if(l+1){ // B数比较大时                 while(l>=0){                    S[j]=(B[l]-'0')+flag;                    if(S[j]>=10){ // 确定进位                         S[j]=S[j]-10;                        flag=1;                    }                    else{                        flag=0;                    }                     l--;                    j++;                }            }             if(flag){ // 确定最大那一位是否会再往前进                 S[j]=flag;            }             else{                j--;            }             printf("Case %d:\n",i+1);            printf("%s + %s = ",A,B);            while(j>-1){                printf("%d",S[j--]);            }              printf("\n");         } } return 0; }