题目描述

I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.

输入

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.

输出

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.

样例输入

2

1 2

112233445566778899 998877665544332211

样例输出

Case 1:

1 + 2 = 3

 

Case 2:

112233445566778899 + 998877665544332211 = 1111111111111111110

 

解题代码：

#include <stdio.h> #include <stdlib.h> #include <string.h> #define MAX 1005 int main() { int t,i,j,k,count=1;//count用来记实例序号 char a[MAX],b[MAX]; int na[MAX],nb[MAX];//数字数组 int sum[MAX],pre;//pre表示前一位 scanf("%d",&t); while(t--){ memset(sum,0,sizeof(sum)); memset(na,0,sizeof(na)); memset(nb,0,sizeof(nb)); scanf("%s%s",a,b); pre=0; int lena = strlen(a); int lenb = strlen(b); //字符串反转且字符串变数字 for(i=0;i<lena;i++) na[lena-1-i] = a[i]-'0'; for(j=0;j<lenb;j++) nb[lenb-1-j] = b[j]-'0'; int lenx=lena>lenb?lena:lenb; //逐位相加（从最低位开始加） for(k=0;k<lenx;k++){ sum[k]=na[k]+nb[k]+pre/10; pre=sum[k]; } while(pre>9){ sum[lenx]=pre/10;//最高位有了进位了 lenx++; pre/=10; } printf("Case %d:\n",count++); printf("%s + %s = ",a,b); for(i=lenx-1;i>=0;i--){ printf("%d",sum[i]);//输出每一位 每一位是sum数组中的每一项对10取余 } printf("\n"); if(t) printf("\n"); } return 0; }