joj 2308 Problems to Solve很巧啊

joj 2308 Problems to Solve很巧啊

    
    
    
    


     
     
     
     
Your teacher has given you some problems to solve. You must first solve problem 0. After solving each problem i, you must either move on to problem i+1 or skip ahead to problem i+2. You are not allowed to skip more than one problem. For example, {0, 2, 3, 5} is a valid order, but {0, 2, 4, 7} is not because the skip from 4 to 7 is too long.

     
     
     
     
You are given N integers, where the ith integer indicates how much you like problem i. The teacher will let you stop solving problems once the range of pleasantness you've encountered reaches a certain threshold. Specifically, you may stop once the difference between the maximum and minimum pleasantness of the problems you've solved is greater than or equal to a integer K. If this never happens, you must solve all the problems.

     
     
     
     
Input
The input contains several test cases. For each case it contains two positive integer N, K (1<=N<=50, 1<=K<=1000), then N integers follow in the next line, the ith integer ( >=1 and <= 1000 ) indicates how much you like problem i.
     
     
     
     


     
     
     
     
Output

     
     
     
     
For each test case, output the minimum number of problems you must solve to satisfy the teacher's requirements.

     
     
     
     
Sample Input
3 2
     
     
     
     
1 2 3
     
     
     
     
5 4
     
     
     
     
1 2 3 4 5

     
     
     
     
Sample Output
2
     
     
     
     
3 
    
    
    
    

很笨但是很实用！
#include<stdio.h>
int a[51];
int main()
{
int n,k;
while(scanf("%d%d",&n,&k)!=EOF)
{
int i,j;
for(i=0;i<n;i++)
scanf("%d",a+i);
int min=0xfffffff;
int flag=0;
for(i=1;i<n;i++)
for(j=0;j<i;j++)
{
if(a[i]-a[j]>=k||a[j]-a[i]>=k)
{
flag=1;
int step;
if(j==0)
{
step=(i+1)/2+1;
if(min>step)
min=step;
}
else
{
step=(i-j+1)/2+(j+1)/2+1;
if(min>step)
min=step;
}
}
}
if(flag)
printf("%d\n",min);
else
printf("%d\n",n);
}
return 0;
}