HDU--1002 -- 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): 165473    Accepted Submission(s): 31614

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

Code:

#include"stdio.h"#include"string.h"int main(){char num1[1001],num2[1002];int n1[1002],n2[1002];int T,ca,i,j,l1,l2,l3;scanf("%d",&T);for(ca=1;ca<=T;ca++){memset(n1,0,sizeof(n1));//全部置为0 memset(n2,0,sizeof(n2));scanf("%s%s",num1,num2);l1 = strlen(num1);l2 = strlen(num2);l3 = l1>l2?l1:l2;printf("Case %d:\n",ca);//输出 for(i=0;i<l1;i++)printf("%c",num1[i]);printf(" + ");for(i=0;i<l2;i++)printf("%c",num2[i]);printf(" = ");for(i=0,j=l1-1;i<l1,j>=0;j--)//字符串逆置 n1[i++] = num1[j] - '0';for(i=0,j=l2-1;i<l2,j>=0;j--)n2[i++] = num2[j] - '0';for(i=0;i<1001;i++)//相加 {n1[i] = n1[i] + n2[i];if(n1[i]>9){n1[i] = n1[i] - 10;n1[i+1]++;}}if(n1[l3]) l3++;//若最高位有进位,则输出个数加1 <l3为最大长度,有进位说明长度加1>for(i=l3-1;(i>=0 && n1[i]=='0');i--);if(i>=0)for(;i>=0;i--)printf("%d",n1[i]);else printf("0");//0+0时空循环中的i变为-1,需另考虑 printf("\n");if(ca!=T)printf("\n");}}

要比较长度啊!! 啊啊啊啊