【codeforces】- Petya and Java（给不同的数，判断大小，选择不同数据类型）

A. Petya and Java

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Little Petya has recently started attending a programming club. Naturally he is facing the problem of choosing a programming language. After long considerations he realized that Java is the best choice. The main argument in favor of choosing Java was that it
has a very large integer data type, called BigInteger.

But having attended several classes of the club, Petya realized that not all tasks require using the BigInteger type. It turned out that in some tasks it is much easier to use small data types. That's why a question arises: "Which integer type to use if one
wants to store a positive integer n?"

Petya knows only 5 integer types:

1) byte occupies 1 byte and allows you to store numbers from  - 128 to 127
2) short occupies 2 bytes and allows you to store numbers from  - 32768 to 32767

3) int occupies 4 bytes and allows you to store numbers from  - 2147483648 to 2147483647

4) long occupies 8 bytes and allows you to store numbers from  - 9223372036854775808 to9223372036854775807

5) BigInteger can store any integer number, but at that it is not a primitive type, and operations with it are much slower.

For all the types given above the boundary values are included in the value range.

From this list, Petya wants to choose the smallest type that can store a positive integer n. Since BigInteger works much slower, Peter
regards it last. Help him.

Input

The first line contains a positive number n. It consists of no more than 100 digits
and doesn't contain any leading zeros. The number n can't be represented as an empty string.

Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to usecout (also
you may use %I64d).

Output

Print the first type from the list "byte, short, int, long, BigInteger",
that can store the natural number n, in accordance with the data given above.

Examples

input

127

output

byte

input

130

output

short

input

123456789101112131415161718192021222324

output

BigInteger

用字符串把大数单独拿出来判断，其他的就可以直接和临界值比较判断类型。有一点，是负数是要注意位数

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int main()
{
char a[111];
int l;
while (~scanf("%s",a))
{
l = strlen(a);
if (a[0]!='-')
{
if (l>19)
printf("BigInteger\n");
else if(l==19)
{
int t=strcmp(a,"9223372036854775807");
if(t==-1||t==0)
printf("long\n");
else
printf("BigInteger\n");
}
else if(l>=11&&l<19)
printf("long\n");
else
{
__int64 n=a[0]-'0';
for (int i=1;i<l;i++)
n=n*10+(a[i]-'0');
if (n>=0&&n<=127)
printf("byte\n");
else if(n>=128&&n<=32767)
printf("short\n");
else if(n>=32768&&n<=2147483647)
printf("int\n");
else
printf("long\n");
}
}
else
{
l--;
if (l>19)
printf("BigInteger\n");
else if(l>=11&&l<19)
printf("long\n");
else if(l==19)
{
int t=strcmp(a,"-9223372036854775808");
if(t==-1||t==0)
printf("long\n");
else
printf("BigInteger\n");
}
else
{
__int64 n=a[1]-'0';
for(int i=2;i<=l;i++)
n=n*10+(a[i]-'0');
if(n>=0&&n<=128)
printf("byte\n");
else if(n>=129&&n<=32768)
printf("short\n");
else if(n>=32769&&n<=2147483648)
printf("int\n");
else
printf("long\n");
}
}
}
return 0;
}