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): 430165 Accepted Submission(s): 83626
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>
long int max(long int c,long int d)
{
if(c>d)
return c;
else
return d;
}
int main()
{
long int n,m,i,t,x,y,j=0;
char a[10000],b[10000],c[10001];
scanf("%ld",&n);
while(n--)
{
scanf("%s",a);
scanf("%s",b);
j++;
x=strlen(a);
y=strlen(b);
t=max(x,y);
for(i=0;i<=t;i++)
c[i]='0';
for(i=0;i<t;i++)
{
if((x-1-i)>=0&&(y-1-i)>=0)
m=a[x-1-i]+b[y-1-i]-96;
else
{
if((x-1-i)<0&&(y-1-i)>=0)
m=b[y-1-i]-48;
else
{
if((x-1-i)>=0&&(y-1-i)<0)
m=a[x-1-i]-48;
}
}
if(m>=10)
{
c[t-1-i]=c[t-1-i]+1;
c[t-i]=c[t-i]+(m-10);
}
else
{
if((c[t-i]-48+m)>=10)
{
c[t-1-i]=c[t-1-i]+1;
c[t-i]=c[t-i]+(m-10);
}
else
{
c[t-i]=c[t-i]+m;
}
}
}
printf("Case %d:\n",j);
printf("%s + %s = ",a,b);
if(c[0]!='0')
printf("%c",c[0]);
for(i=1;i<=t;i++)
{
printf("%c",c[i]);
}
printf("\n");
if(n!=0)
{
printf("\n");
}
}
return 0;
}