int main ()
{
int n, i, j, cnt, a[510] ;
while (~scanf ("%d",&n))
{
for(i = 0 ; i < n ; i++)
scanf ("%d", &a[i]) ;
cnt = 0 ;
for (i = n-1 ; i > 0 ; i--)
for (j = 0 ; j < i; j++) if (a[j]>a[j+1])
{
cnt++ ;
a[j]^= a[j+1]^=a[j]^=a[j+1];
}
printf ("%d\n", cnt) ;
}
return 0 ;
}
# include <stdio.h>
# include <string.h>
int vis[510], a[510] ;
int main ()
{
int n, m, k ;
int i, j, cnt, tmp, p, q ;
while (~scanf ("%d",&n))
{
for(i = 1 ; i <= n ; i++)
scanf ("%d", &a[i]) ;
cnt = 0 ;
for (i = 1 ; i <= n ; i++)
{
while (a[i] != i)
{
cnt++ ;
p = a[i], q = i ;
a
[原创]C语言初级教学之百钱买百鸡的循环实现
赞一个,楼主加油~也可以去OJ上做做题。
晒晒mathematica源码:
Solve[{a + b + c == 100, 5 a + 3 b + c/3 == 100, a >= 0, b >= 0, c >= 0}, {a, b, c}, Integers]
结果
{{a -> 0, b -> 25, c -> 75}, {a -> 4, b -> 18, c -> 78}, {a -> 8,
b -> 11, c -> 81}, {a -> 12, b -> 4, c -> 84}}