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 integers a1,?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.
Examples
Input
4 1 1 2 1 5
Output
2 + 3 2 - 4 1
Input
4 1 1 2 3 4
Output
0
本题的意思就是首先输入两个数n，k表示第二行将要输入的数组长度以及要求改变该数组为公差为k的等差数列；从题目中我们可以得到公差最大为1000；那么给我们一串数据例如
4 1
1 3 4 5
这时候我们输出的应当是
1
+ 1 1
这时候我们发现，我们需要找到适合的修改区间，也就是需要修改最少的那一段，AC代码奉上
#include<bits/stdc++.h>//万能头yyds，但是删除了就啥也不是，如果竞赛省时间建议使用，私底下练习还是老老实实记住头文件吧
using namespace std;
int a[1001],b[1001];
int main() {int n,k;int i,j;cin>>n>>k; memset(b,0,sizeof(b));for(i=0; i<n; i++)cin>>a[i];//录入数据 for(i=1; i<1001; i++) {//所有数据比对变大，找到适合的段，也就是需要改变最少的段 for(j=0; j<n; j++) {if(a[j]==i+j*k)b[i]++;//记录不需要改变的树木 }}int y=b[0];int f=0;for(i=1; i<1001; i++)if(y<b[i]) {y=b[i];f=i;}cout<<n-y<<endl;//总的树木减去不需要改变的树木 for(i=0; i<n; i++) {if(a[i]<f+i*k)cout<<"+ "<<i+1<<" "<<f+i*k-a[i]<<endl;if(a[i]>f+i*k)cout<<"- "<<i+1<<" "<<a[i]-f-i*k<<endl;}return 0;
}