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            posts - 195,  comments - 30,  trackbacks - 0

            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
            很笨但是很實用!
            #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;
            }
            posted on 2009-06-28 22:26 luis 閱讀(249) 評論(0)  編輯 收藏 引用 所屬分類: 給我啟發題
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