A + B Problem II

时间：2015-06-17 15:25:27 阅读：148 评论：0 收藏：0 [点我收藏+]

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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 256935 Accepted Submission(s): 49653

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>
int main()
{
char a[1000],b[1000],c[1003];
int i,j=1,p,n,n1,n2;
scanf("%d",&n);
while(n)
{
scanf("%s %s",a,b);
n1=strlen(a)-1;
n2=strlen(b)-1;
for(i=0;n1>=0||n2>=0;i++,n1--,n2--)
{
p=0;
if(n1>=0&&n2>=0)
c[i]=a[n1]+b[n2]-‘0‘+p;
if(n1>=0&&n2<0)
c[i]=a[n1]+p;
if(n1<0&&n2>=0)
c[i]=b[n2]+p;
if(c[i]>‘9‘)
{
c[i]=c[i]-10;
p=1;
}
}
if(p==1)printf("%d",p);
printf("Case %d:\n",j);
printf("%s + %s = ",a,b);
while(i--)
printf("%c",c[i]);
j++;
if(n!=1)
printf("\n\n");
else
printf("\n");
n--;
}
return 0;
}

A + B Problem II

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原文地址：http://blog.csdn.net/l15738519366/article/details/46533257

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