# include & ltstring.h & gt
# DefinitionNo. 1005
int main()
{
char str 1[N]，str 2[N]；
int num 1[N]，num2[N]，t，len 1，len2，I，j，max
scanf("%d "，& ampt)； //Test T data, and don't write subtraction.
getchar()；
while(t -)
{
memset(str 1，0，sizeof(str 1))；
memset(str 1，0，sizeof(str 2))；
memset(num 1，0，sizeof(num 1))；
memset(num2，0，sizeof(num 2))；
scanf("%s "，str 1)； //Enter the first number. This 1000 bit integer can be added together. You can change its length yourself.
scanf("%s "，str 2)； //Enter the second number, because there is no long data type, and use a string to represent a large integer.
len 1 = strlen(str 1)；
len 2 = strlen(str 2)；
max = len 1 & gt； len2？ len 1:len 2；
j = 0；
for(I = len 1- 1； I & gt=0; I-)// Note here that the conversion is reversed, but num 1 stores each bit in the forward direction.
{
num 1[j]= str 1[I]--0； //Converted into numbers and stored in the array one by one.
j++；
}
j = 0；
for(I = len 2- 1； I & gt=0; I-)
{
num 2[j]= str 2[I]--0；
j++；
}
for(I = 0； I & ltmaxI++) // Add it bit by bit.
{
num 2[I]+= num 1[I]； //Add to num2 array
if(num 2[I]& gt； = 10) // Decimal greater than 10.
{
num 2[I]-= 10；
num 2[I+ 1]+= 1；
}
}
if(num2[max]== 1)
printf("%d "，num 2[max])；
for(I = max- 1； I & gt=0; I-)// In this way, every bit of a big integer is calculated, so just output it in order.
printf("%d "，num 2[I])；
printf(" \ n ")；
}
Returns 0;
}