我刚刚也写了一个类似的小程序,时间有限可能不是很好,你可以试试。
void main()
{
int a[5];
int x;
printf( "输入5个数:\n" );
for( int i = 0; i < 5; ++i )
scanf( "%d", a[i] );
for( int i = 5; i >2; --i )
{
if( 5 == i )
{
for( m = 0; m < 5; ++m )
for( int n = 0; n < 5; ++n )
if( a[m] < a[n] )
{
x = a[m];
a[m] = a[n]
a[n] = x;
}
for( int p = 0; p < i; ++p )
printf( "%d", a[p]);
printf( "\n" );
}
if( 4 == i)
{
for( int j = 0; j < 5; ++j)
{
for( m = 0; m < 5; ++m )
if( m != j)
for( int n = 0; n < 5; ++n )
if( n != j )
if( a[m] > a[n] )
{
x = a[m];
a[m] = a[n];
a[n] = x;
}
for( int p = 0; p < i; ++p )
printf( "%d", a[p] );
printf( "\n" );
}
}
if( 3 == i )
{
for( int j = 0; j < 5; ++j )
for( int k = j+1; k < 5; ++k )
{
for( int m = 0; m < 5; ++m )
if( ( m != j ) && ( m != k ) )
for( int n = 0; n < 5; ++n )
if( ( n != j ) && ( n != k ) )
if( a[m] < a[n] )
{
x = a[m];
a[m] = a[n];
a[n] = x;
}
for( int p = 0; p < i; ++p )
printf( "%d", a[p] );
printf( "\n" );
}
}
}
}
#include
void main()
{
int year,leap;
printf("请输入一个年数:\n");
scanf("%d",&year);
if(year%4==0)
{
if(year%100==0)
{
if(year%400==0) leap=1;//将==改成=
else leap=0;//将==改成=
}
else leap=1;
}
else
leap=0;//将==改成=
if(leap)
printf("%d is",year);
else
printf("%d is not",year);
printf("a leap year.\n");
}
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