codeforces 402B Trees in a Row-优快云博客

本文介绍了一个有趣的问题：如何通过增加或减少树的高度来使花园中的一排树形成等差数列，同时确保所需操作次数最少。文章提供了一种有效的算法解决方案，并附带了实现代码。

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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 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.

Sample test(s)

Input

4 11 2 1 5

Output

2+ 3 2- 4 1

Input

4 11 2 3 4

Output

直接暴力就行，需要注意树高不能小于0，害我错了好长时间；

代码:

#include <iostream>
#include <math.h>
using namespace std;

int main()
{
    int n,k;
    int a[1010];
   // int num[1010];
    while(cin>>n>>k)
    {
        for(int i=1;i<=n;i++)
        {
            cin>>a[i];
        }
        int maxn=1;
        int cnt=0;
        int ans=1;

        for(int i=1;i<=n;i++)
        {   cnt=0;
           int flag=1;
            for(int j=1;j<=n;j++)
            {
                if((j-i)*k+a[i]<=0) {flag=0;break;}
                if(a[j]==(j-i)*k+a[i])
                {
                    cnt++;
                }
            }
            if(!flag) continue;
            if(maxn<cnt)
            {
                ans=i;
                maxn=cnt;
            }
        }
        //cout<<ans<<endl;
        cout<<n-maxn<<endl;
        for(int i=1;i<=n;i++)
        {
            if(a[i]<(i-ans)*k+a[ans])
            {
                cout<<"+"<<" "<<i<<" "<<fabs((i-ans)*k+a[ans]-a[i])<<endl;
            }
            else if(a[i]>(i-ans)*k+a[ans])
            cout<<"-"<<" "<<i<<" "<<fabs((i-ans)*k+a[ans]-a[i])<<endl;
        }
    }
    return 0;
}