数据的交换输出
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 54693 Accepted Submission(s): 20777
4 2 1 3 4 5 5 4 3 2 1 0
1 2 3 4 1 4 3 2 5
#include<stdio.h>
#include<math.h>
int main()
{
int n,i,m,p;
int a[110];
while(scanf("%d",&n)!=EOF&&(n!=0))
{
scanf("%d",&a[0]);
m=0;
for(i=1;i<n;i++)
{
scanf("%d",&a[i]);
if(a[m]>a[i])
m=i;
}
p=a[0];
a[0]=a[m];
a[m]=p;
for(i=0;i<n;i++)
{
if(i!=n-1)
printf("%d ",a[i]);
else
printf("%d\n",a[i]);
}
}
return 0;
}
//怎样找到最小数与最前面数交换,其他数不改变顺序输出