A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 260647 Accepted Submission(s): 50397
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 as 1000
int main()
{
int as1[as+],as2[as+],cot=,j,t;
char shuru1[as+],shuru2[as+];
int n,i;
scanf("%d",&n);
getchar();//吸收回车符
t=n;
while(n--)
{
cot++;
scanf("%s",shuru1);
scanf("%s",shuru2);
memset(as1,,sizeof(as1));
memset(as2,,sizeof(as2));
for(i=,j=strlen(shuru1)-;j>=;j--,i++)
{
as1[i]=shuru1[j]-''; }
for(i=,j=strlen(shuru2)-;j>=;j--,i++)
{
as2[i]=shuru2[j]-'';
}
for(i=;i<as+;i++)
{
as1[i]+=as2[i];
if(as1[i]>=)//判断是否满十进一
{
as1[i+]++;
as1[i]-=;
} }
for(i=as+;(i>=)&&(as1[i]==);i--);//去掉结果前面多余的0
printf("Case %d:\n",cot);
printf("%s + %s = ",shuru1,shuru2);
if(i>=)
{
for(;i>=;i--)
printf("%d",as1[i]);
}
else
printf("");
printf("\n");
if(cot!=t)
printf("\n");//最后不要多换行
}
return ;
}