Codeforces Round #236 (Div. 2)B - c++编程基础

本题枚举第一课树的高度即可

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 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 integersa1, 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.

Sample test(s) input

4 1
1 2 1 5

output

2
+ 3 2
- 4 1

input

4 1
1 2 3 4

output

#include 
  
   
#include
   
     #define MAX 1001 int k,n,hight[MAX],f[MAX],ans[MAX]; int main() { //freopen("input.txt","r",stdin); while(scanf("%d%d",&n,&k)!=EOF) { scanf("%d",hight+1); for(int i=2;i<=n;i++)scanf("%d",hight+i); int min=1000,p=hight[1]; memset(f,0,sizeof(f)); for(int i=1;i<=1000;i++) { int d=1; if(i==p)d=0; int temp1=hight[1]; f[1]=i-hight[1]; for(int j=2;j<=n;j++) { temp1=temp1+k; f[j]=temp1-hight[j]; if(f[j]!=0)d++; } if(d
    
     0)printf("+ %d %d\n",i,ans[i]); else printf("- %d %d\n",i,ans[i]); } } return 0; }