Codeforces Round #236 (Div. 2)B

本题枚举第一课树的高度即可

B. Trees in a Row

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

The Queen of England has n trees growing in a row in her garden. At that, the i-th (1 ≤ i ≤ n) tree from the left has height ai meters. Today the Queen decided to update the scenery of her garden. She wants the trees' heights to meet the condition: for all i (1 ≤ i < n),ai + 1 - ai = k, where k is the number the Queen chose.

Unfortunately, the royal gardener is not a machine and he cannot fulfill the desire of the Queen instantly! In one minute, the gardener can either decrease the height of a tree to any positive integer height or increase the height of a tree to any positive integer height. How should the royal gardener act to fulfill a whim of Her Majesty in the minimum number of minutes?

Input

The first line contains two space-separated integers: n, k (1 ≤ n, k ≤ 1000). The second line contains n space-separated integersa1, a2, ..., an (1 ≤ ai ≤ 1000) — the heights of the trees in the row.

Output

In the first line print a single integer p — the minimum number of minutes the gardener needs. In the next p lines print the description of his actions.

If the gardener needs to increase the height of the j-th (1 ≤ j ≤ n) tree from the left by x (x ≥ 1) meters, then print in the corresponding line "+ j x". If the gardener needs to decrease the height of the j-th (1 ≤ j ≤ n) tree from the left by x (x ≥ 1) meters, print on the corresponding line "- j x".

If there are multiple ways to make a row of trees beautiful in the minimum number of actions, you are allowed to print any of them.

Sample test(s)

input

4 1
1 2 1 5

output

2
+ 3 2
- 4 1

input

4 1
1 2 3 4

output

0

#include <cstdio>
#define MAX 1001
int k,n,hight[MAX];
int main()
{
   //freopen("input.txt","r",stdin);
   while(scanf("%d %d",&n,&k)!=EOF)
   {
      for(int i=1;i<=n;i++)scanf("%d",hight+i);
      int min=1001,temp=0;
      for(int i=1;i<=1000;i++)
      {
          int d=0;
          for(int j=1;j<=n;j++)
          {
              if((i+(j-1)*k)!=hight[j])d++;
          }
          if(d<min)
          {
              min=d;
              temp=i;
          }
      }
      printf("%d\n",min);
      for(int i=1;i<=n;i++)
      {
          int f=temp+(i-1)*k-hight[i];
          if(f<0)printf("- %d %d\n",i,-f);
          if(f>0)printf("+ %d %d\n",i,f);
      }
   }
    return 0;
}