本文主要是介绍Codeforces Round #236 (Div. 2)B，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
B. 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 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?
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.
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.
4 1 1 2 1 5
2 + 3 2 - 4 1
4 1 1 2 3 4
0
#include <cstdio>
#define MAX 1001
int k,n,hight[MAX];
int main()
{//freopen("input.txt","r",stdin);while(scanf("%d %d",&n,&k)!=EOF){for(int i=1;i<=n;i++)scanf("%d",hight+i);int min=1001,temp=0;for(int i=1;i<=1000;i++){int d=0;for(int j=1;j<=n;j++){if((i+(j-1)*k)!=hight[j])d++;}if(d<min){min=d;temp=i;}}printf("%d\n",min);for(int i=1;i<=n;i++){int f=temp+(i-1)*k-hight[i];if(f<0)printf("- %d %d\n",i,-f);if(f>0)printf("+ %d %d\n",i,f);}}return 0;
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