10055 - Hashmat the Brave Warrior & 各数据类型所占字节数（C语言）

浏览数：49 / 时间：2015年06月08日

Problem A

Hashmat the brave warrior

Input: standard input

Output: standard output

Hashmat is a brave warrior who with his group of young soldiers moves from one place to another to fight against his opponents. Before fighting he just calculates one thing, the difference between his soldier number and the opponent‘s soldier number. From this difference he decides whether to fight or not. Hashmat‘s soldier number is never greater than his opponent.

Input

The input contains two integer numbers in every line. These two numbers in each line denotes the number of soldiers in Hashmat‘s army and his opponent‘s army or vice versa. The input numbers are not greater than 2^32. Input is terminated by End of File.

Output

For each line of input, print the difference of number of soldiers between Hashmat‘s army and his opponent‘s army. Each output should be in seperate line.

Sample Input:

10 12
10 14
100 200

Sample Output:

2
4
100

______________________________________________

题目解答:

 1 #include<stdio.h>
 2 int main(){
 3     long a,b,c; 
 4     while(scanf("%ld%ld",&a,&b) == 2){
 5         c = a - b;
 6         if ( c < 0) c = -c;
 7         printf("%ld\n",c);
 8     }
 9 
10     return 0;   
11 }

注意点：

1.　　输入的数据不大于2^32,32位系统或者是64位使用的int的表示范围为(-2^31)~(2^31-1)。使用long

2.　　使用long可以accept，UVa平台使用的是64位的。在32位系统中long也是占4字节，不满足题目要求。

获取各数据类型所占字节数：我的电脑为64位的系统上:(使用sizeof(int))

#include<stdio.h>

int main(){
    printf("short:%d\n",sizeof(short));
    printf("signed int:%d\n",sizeof(signed int));
    printf("unsigned int:%d\n",sizeof(unsigned int));
    printf("long:%d\n",sizeof(long));
    printf("long long:%d\n",sizeof(long long));
    printf("double:%d\n",sizeof(double));
    printf("float:%d\n",sizeof(float));

    return 0;
    
}