codeforces402B

Trees in a Row

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 integers a₁, 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.

Examples

Input

4 1
1 2 1 5

Output

2
+ 3 2
- 4 1

Input

4 1
1 2 3 4

Output

0

sol:数据范围小的可怜，爆枚一个正确节点，n²模拟即可

#include <bits/stdc++.h>

using namespace std;

typedef int ll;

inline ll read()

{

    ll s=;

    bool f=;

    char ch=' ';

    while(!isdigit(ch))

    {

        f|=(ch=='-'); ch=getchar();

    }

    while(isdigit(ch))

    {

        s=(s<<)+(s<<)+(ch^); ch=getchar();

    }

    return (f)?(-s):(s);

}

#define R(x) x=read()

inline void write(ll x)

{

    if(x<)

    {

        putchar('-'); x=-x;

    }

    if(x<)

    {

        putchar(x+''); return;

    }

    write(x/);

    putchar((x%)+'');

    return;

}

#define W(x) write(x),putchar(' ')

#define Wl(x) write(x),putchar('\n')

const int N=;

int n,m,a[N],b[N],Ans[N];

int main()

{

    int i,j,Pos=-;

    R(n); R(m);

    for(i=;i<=n;i++) R(a[i]);

    for(i=;i<=n;i++)

    {

        Ans[i]=;

        b[i]=a[i];

        for(j=i-;j>=;j--) b[j]=b[j+]-m;

        for(j=i+;j<=n;j++) b[j]=b[j-]+m;

        for(j=;j<=n;j++)

        {

            if(b[j]!=a[j]) Ans[i]++;

            if(b[j]<=) {Ans[i]=0x3f3f3f3f; break;}

        }

        if((Pos==-)||(Ans[i]<Ans[Pos])) Pos=i;

    }

    Wl(Ans[Pos]);

    b[Pos]=a[Pos];

    for(i=Pos-;i>=;i--) b[i]=b[i+]-m;

    for(i=Pos+;i<=n;i++) b[i]=b[i-]+m;

    for(i=;i<=n;i++) if(a[i]!=b[i])

    {

        if(a[i]<b[i])

        {

            putchar('+'); putchar(' '); W(i); Wl(b[i]-a[i]);

        }

        else

        {

            putchar('-'); putchar(' '); W(i); Wl(a[i]-b[i]);

        }

    }

    return ;

}

/*

Input

4 1

1 2 1 5

Output

2

+ 3 2

- 4 1

Input

4 1

1 2 3 4

Output

0

*/

巴特西

codeforces402B

Trees in a Row

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