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 解题方法：暴力求解，将分别从1,2,3,4,5……开始的等差数列与所给数据比较

#include 
        
         #include 
         
          using namespace std;int a[1005],b[1005];int main(){    int n,k;    int i,j;    while(cin>>n>>k)    {        memset(b,0,sizeof(b));        for(i=0;i
          
           >a[i];        for(i=1;i<1005;i++)        {            for(j=0;j
           
            f+i*k)            cout<<"- "<
            
             <<" "<
             
              <
             
            
           
          
         
        

 

错误思想：将输入数据分别减去（i-1)*k,如果输入数据是等差数列，这时每个数应该相等（实际输入不是等差数列），从中找出相同数最多的数，其它数则为应当增减的数，此方法存在漏洞：若输入数据为5个，d=1，输入1 1 2 3 4，则处理后为0 1 2 3 4，而题中要求数据大于0，故错误！其代码如下：

#include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            
             using namespace std;int a[1005],b[10000000];int main(){    int n,k;    int i,j,t,x,maxn,minn;    while(cin>>n>>k)    {        minn=99999999;        maxn=-99999999;        memset(b,0,sizeof(b));        for(i=0;i
             
              maxn) maxn=a[i]; if(a[i]
              
               =minn;i--) { for(j=0;j
               
                x) cout<<"- "<
                
                 <<" "<
                 
                  <