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=1005;
const int INF=1<<28;
using namespace std;
int a[N],b[N],f[N];
int main()
{
    int n,d;
    while(cin>>n>>d){
        cin>>f[1];
        int flag1 = 1;
        for (int i = 2; i <= n; i++){
            cin>>f[i];
            if (f[i]-f[i-1]!=d)
                flag1 = 0;
        }
        if (flag1){ //原序列为公差为d的等差序列
            puts("0");
            continue;
        }
        int cnt,Min = INF;
        for (int i = 1; i <= n; i++){
            int m = f[i]-d;
            cnt = 0;flag1 = 0;
            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
                if (m <= 0){ //出现小于0的数，则说明该序列不合法
                    flag1 = 1;
                    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+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
                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,-1,sizeof(a));//a[]存储对数据进行增的操作
                memset(b,-1,sizeof(b));//b[]存储对数据进行减的操作
                for (int j = i-1; j >= 1; 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+1; 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 = 1; i <= n; i++){
            if (a[i]!=-1){
                printf("+");
                cout<<" "<<i<<" "<<a[i]<<endl;
            }
            if (b[i]!=-1){
                printf("-");
                cout<<" "<<i<<" "<<b[i]<<endl;
            }
        }
    }
    return 0;
}