CodeForce 66A Petya and Java
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Description
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 to 9223372036854775807
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 ncan'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 use cout (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.
Sample Input
127
byte
130
short
123456789101112131415161718192021222324
BigInteger
题意:输入一个正数判断是什么类型的数。
思路:很明显要用到字符串知识,这个很容易想到,不过刚开始本想把字符转换为数字再进行比较,不过这样特别麻烦,
换个思路,如果利用字符串中比较大小的函数strcmp,明显简单多了。
#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int main(){char s[110],n,m,i,j,k,l;while(scanf("%s",s)!=EOF){l=strlen(s);if(l<3||l==3&&strcmp(s,"127")<=0)printf("byte\n");else if(l<5||l==5&&strcmp(s,"32767")<=0)printf("short\n");else if(l<10||l==10&&strcmp(s,"2147483647")<=0)printf("int\n");else if(l<19||l==19&&strcmp(s,"9223372036854775807")<=0)printf("long\n");else printf("BigInteger\n");}return 0; }
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