334 人解决,373 人已尝试。
386 份提交通过,共有 1309 份提交。
2.0 EMB 奖励。
单点时限: 2.0 sec
内存限制: 256 MB
输入 n(1<n<11) 个整数,将其中最小数与第一个数交换,将其中最大数与最后一个数交换。
定义 3 个函数:
1) input 输入
2) process 处理交换
3) output 输出
//* Specification of input *
void input(int* p, int n);
/* PreCondition:
p points to an array with n integers
PostCondition:
enter and store n integers into an array pointed by p
*/
//* Specification of process *
void process(int* p, int n);
/* PreCondition:
p points to an array with n integers
PostCondition:
swap smallest number with the first number of an array pointed by p,
and swap largest number with the last number
*/
//* Specification of output *
void output(int* p, int n);
/* PreCondition:
p points to an array with n integers
PostCondition:
print each element of an array pointed by p in one line, with one blank between elements.
*/
只需按要求写出函数定义,并使用给定的测试程序测试你所定义函数的正确性。
不要改动测试程序。测试正确后,将测试程序和函数定义一起提交。
/*********/
/ /
/ DON'T MODIFY main function ANYWAY! /
/ /
/*********/
//* Specification of input *
void input(int* p, int n)
/* PreCondition:
p points to an array with n integers
PostCondition:
enter and store n integers into an array pointed by p
*/
{ //TODO: your function definition
}
//* Specification of process *
void process(int* p, int n)
/* PreCondition:
p points to an array with n integers
PostCondition:
swap smallest number with the first number of an array pointed by p,
and swap largest number with the last number
*/
{ //TODO: your function definition
}
//* Specification of output *
void output(int* p, int n)
/* PreCondition:
p points to an array with n integers
PostCondition:
print each element of an array pointed by p in one line,
with one blank between elements.
*/
{ //TODO: your function definition
}
/*********/
int main()
{ int a[N],n;
scanf(“%d”,&n);
//* functions input, process, output are called here ***
input(a,n);
process(a,n);
output(a,n);
//************
return 0;
}
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334 人解决,373 人已尝试。
386 份提交通过,共有 1309 份提交。
2.0 EMB 奖励。
创建: 9 年前.
修改: 7 年,3 月前.
最后提交: 10 月前.
来源: N/A