1002 A + B Problem II

来源：互联网发布：java crc16校验算法编辑：程序博客网时间：2024/05/01 21:35

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 224674 Accepted Submission(s): 43094

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

注意：0+0=0  此题有点坑啊，002+003=005竟然可以通过，我也是醉了

#include<cstdio>#include<string.h>#include<algorithm>#include<iostream>using namespace std;char a[1000],b[1000];int a1[1000],b1[1000],c[1005],MAX;void zhuan(int m,int n){ //字符转化为整数序列    int j=0,z=0;   for(int i=m-1; i >= 0; i--)a1[j++]=a[i]-48;           for(int i=n-1; i >= 0; i--)b1[z++]=b[i]-48;}    void find(int n){      for(int i=0; i< n;i++){            int s = a1[i] + b1[i] + c[i];            c[i]=s%10;            c[i+1]=s/10;                    }      if(c[n])MAX++;}int main(){    int test,count=0;    scanf("%d",&test);    int N = test;    while(test--){                  scanf("%s%s",a,b);      memset(c,0,sizeof(c));      memset(a1,0,sizeof(a1));      memset(b1,0,sizeof(b1));      int m = strlen(a);      int n = strlen(b);      zhuan(m,n);      MAX=m >= n ? m : n;      find(MAX);      printf("Case %d:\n",++count);      printf("%s + %s = ",a,b);      for(int i=MAX-1;i>=0;i--)//输出运算结果             {                 printf("%d",c[i]);             }       printf("\n");      if(count != N)           printf("\n");                                                      }                return 0; }

0 0