B. Trees in a Row(cf)

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 a_i 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),a_i + 1 - a_i = 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, ..., a_n (1 ≤ a_i ≤ 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

题意：对一个序列，通过最少次的对某些数据的增或减，使得该序列成为公差为k的等差序列。

思路：遍历所有的数，假设当前数为等差序列中的数，通过增减其它数，将所有的数变为等差序列中的数，记录最小的操作次数，并将操作的数记录下来。注意：调整后序列中的值必须全部大于0，在这跪了好多次。。。

 #include <stdio.h>
 #include <string.h>
 #include <algorithm>
 #include <iostream>
 const int N=;
 const int INF=<<;
 using namespace std;
 int a[N],b[N],f[N];
 int main()
 {
     int n,d;
     while(cin>>n>>d){
         cin>>f[];
         int flag1 = ;
         for (int i = ; i <= n; i++){
             cin>>f[i];
             if (f[i]-f[i-]!=d)
                 flag1 = ;
         }
         if (flag1){ //原序列为公差为d的等差序列
             puts("");
             continue;
         }
         int cnt,Min = INF;
         for (int i = ; i <= n; i++){
             int m = f[i]-d;
             cnt = ;flag1 = ;
             for (int j = i-; j >= ; j--){//调整f[i]左边的数使其成为公差为d的等差序列
                 if (m <= ){ //出现小于0的数，则说明该序列不合法
                     flag1 = ;
                     break;
                 }
                 if (f[j]!=m)
                     cnt++;//记录修改的数的个数
                 m = m-d;
             }
             if (flag1)
                 continue;
             m = f[i]+d;
             for (int j = i+; j <= n; j++){//调整f[i]右边的数使其成为公差为d的等差序列
                 if (f[j]!=m)
                     cnt++;
                     m+=d;
             }
             if (cnt < Min){
                 Min = cnt;//记录最少的操作次数
                 m = f[i]-d;
                 memset(a,-,sizeof(a));//a[]存储对数据进行增的操作
                 memset(b,-,sizeof(b));//b[]存储对数据进行减的操作
                 for (int j = i-; j >= ; j--){
                     if (f[j] > m)
                         b[j] = f[j]-m;
                     if (f[j] < m)
                         a[j] = m-f[j];
                         m-=d;
                 }
                 m = f[i]+d;
                 for (int j = i+; j <= n; j++){
                     if (f[j] > m)
                         b[j] = f[j]-m;
                     if (f[j] < m)
                         a[j] = m-f[j];
                     m+=d;
                 }
             }
         }
         cout<<Min<<endl;
         for (int i = ; i <= n; i++){
             if (a[i]!=-){
                 printf("+");
                 cout<<" "<<i<<" "<<a[i]<<endl;
             }
             if (b[i]!=-){
                 printf("-");
                 cout<<" "<<i<<" "<<b[i]<<endl;
             }
         }
     }
     return ;
 }