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。。。

1 #include 
       
         2 #include 
        
          3 #include 
         
           4 #include 
          
            5 const int N=1005; 6 const int INF=1<<28; 7 using namespace std; 8 int a[N],b[N],f[N]; 9 int main()10 {11     int n,d;12     while(cin>>n>>d){13         cin>>f[1];14         int flag1 = 1;15         for (int i = 2; i <= n; i++){16             cin>>f[i];17             if (f[i]-f[i-1]!=d)18                 flag1 = 0;19         }20         if (flag1){ //原序列为公差为d的等差序列21             puts("0");22             continue;23         }24         int cnt,Min = INF;25         for (int i = 1; i <= n; i++){26             int m = f[i]-d;27             cnt = 0;flag1 = 0;28             for (int j = i-1; j >= 1; j--){
     //调整f[i]左边的数使其成为公差为d的等差序列29                 if (m <= 0){ //出现小于0的数，则说明该序列不合法30                     flag1 = 1;31                     break;32                 }33                 if (f[j]!=m)34                     cnt++;//记录修改的数的个数35                 m = m-d;36             }37             if (flag1)38                 continue;39             m = f[i]+d;40             for (int j = i+1; j <= n; j++){
     //调整f[i]右边的数使其成为公差为d的等差序列41                 if (f[j]!=m)42                     cnt++;43                     m+=d;44             }45             if (cnt < Min){46                 Min = cnt;//记录最少的操作次数47                 m = f[i]-d;48                 memset(a,-1,sizeof(a));//a[]存储对数据进行增的操作49                 memset(b,-1,sizeof(b));//b[]存储对数据进行减的操作50                 for (int j = i-1; j >= 1; j--){51                     if (f[j] > m)52                         b[j] = f[j]-m;53                     if (f[j] < m)54                         a[j] = m-f[j];55                         m-=d;56                 }57                 m = f[i]+d;58                 for (int j = i+1; j <= n; j++){59                     if (f[j] > m)60                         b[j] = f[j]-m;61                     if (f[j] < m)62                         a[j] = m-f[j];63                     m+=d;64                 }65             }66         }67         cout<
           
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