输入一个正整数n(1<n<=10),再输入n个整数存入数组a中,先将数组a中的n个 C语言,C语言! 输入一个正整数n(1<n<=10),再输入...

\u8f93\u5165\u4e00\u4e2a\u6b63\u6574\u6570n (1<n\u226410),\u518d\u8f93\u5165n\u4e2a\u6574\u6570,\u5c06\u5b83\u4eec\u5b58\u5165\u6570\u7ec4a\u4e2d\u3002

#include
#include

int main(void)
{
int n = 0, i;
int min = 0,minpos = 0;
int* a = NULL;
printf("\u8bf7\u8f93\u5165\u4f60\u8981\u8f93\u5165\u591a\u5c11\u6574\u6570\n");
scanf("%d",&n);

a = (int*)malloc(n*sizeof(int));

if(a == NULL) return -1;

printf("\u8bf7\u8f93\u5165\u8fd9%d\u4e2a\u6574\u6570\n",n);
for(i=0; i<n;)
{
scanf("%d",a+i);
if(*(a+i)10)
{
printf("\u91cd\u65b0\u8f93\u5165\n");
}else{
i++;
}
}

min = *a;
for(i=0; i<n; i++)
{
if(*(a+i) < min)
{
min = *(a+i);
minpos = i;
}

}

printf("\u6700\u5c0f\u6570%d,\u4f4d\u7f6e%d\n",min, minpos+1);

*(a+minpos) = *a;
*a = min;

for(i=0; i<n; i++)
{
printf("%d\t",*(a+i));
}
return 0;
}

//\u5f53\u8f93\u5165\u7684\u6570\u53ef\u4ee5\u91cd\u590d\u7684\u65f6\u5019\uff0c\u6682\u65f6\u6ca1\u8003\u8651\u8fd9\u4e2a\uff0c\u9700\u8981\u7684\u8bdd\uff0c\u76f4\u63a5\u4fee\u6539\u5b58\u50a8\u53d8\u91cf\u7684\u4ee3\u7801\u5373\u53ef

#include void main(){int i,n,k=0,max,a[10];printf("\u8f93\u5165\u6b63\u6574\u6570n(n\u5927\u4e8e1\u4e14\u4e0d\u5927\u4e8e10): ");scanf("%d",&n);printf("\u8f93\u5165%d\u4e2a\u6574\u6570: ",n);scanf("%d",&a[0]);max=a[0];for(i=1;i<n;i++){scanf("%d",&a[i]);if(max<a[i]){max=a[i];k=i;}}printf("\u6700\u5927\u503c: %d\n\u5bf9\u5e94\u7684\u4e0b\u6807: %d",max,k);}

#include <stdio.h>

int main()

{

int temp;

int a[],n;

for(int i=0;i<n;i++)

scanf("%d",&a[i]);

for(int i=0;i<n/2;i++){

temp=a[i];

a[i]=a[9-i];

a[9-i]=temp;

}

for(int i=0;i<n;i++)

printf("%d  ",a[i]);  

return 0;

}

扩展资料:

C语言书写规则:

1、一个说明或一个语句占一行。

2、用{} 括起来的部分,通常表示了程序的某一层次结构。{}一般与该结构语句的第一个字母对齐,并单独占一行。

3、低一层次的语句或说明可比高一层次的语句或说明缩进若干格后书写。以便看起来更加清晰,增加程序的可读性。在编程时应力求遵循这些规则,以养成良好的编程风格。

参考资料:

百度百科-c语言



#include <stdio.h>
int main(void){
    int a[10],n,i;
    printf("Input n(int 1<n<=10)...
n=");
    if(scanf("%d",&n)!=1 || n<2 || n>10){
        printf("Input error, exit...
");
        return 0;
    }
    for(i=n-1;i>=0;scanf("%d",a+i--));//n个数逆序存放
    for(i=0;i<n;printf("%d ",a[i++]));//看一看是不是逆序存放的
    printf("
");
    for(i=n-1;i>=0;printf("%d ",a[i--]));//现在正序输出
    printf("
");
    return 0;
}


#include<stdio.h>
int main()
{
int a[10],i,j,t,n;
printf("input n\n");
scanf("%d",&n); // 输入 n
if (n>10) n=10; //修正 n
if (n<=1) n=2; //修正 n
printf("input %d data:\n",n);
for (i=0;i<n;i++) scanf("%d",&a[i]); //输入数据
for (i=0;i<n/2;i++) {t=a[i];a[i]=a[n-1-i];a[n-1-i]=t;}; //逆序
for (i=0;i<n;i++) printf("%d ",a[i]); // 输出
return 0;
}

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