Trees in a Row（暴力）

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

0

 

    题意：

    给出 N，K（1 ~ 1000）。代表有N个数，输出要如何操作最少数，使整个序列满足 ai - ai-1 = k。输出最少操作数，后输出这些操作是什么。（操作  + 序号 + 个数）。每个数都必须 >= 0。

 

    思路：

    暴力。1000个数，每个数都扫一遍也就O（n^2）< 1s，果断暴搜。每个位置都向两边搜索一遍，同时比较最小操作数，若有数小于等于0则不取这个点。一开始没看到所有数都必须 >= 0，无限wa。 

 

    AC：

#include <stdio.h>
#include <string.h>

int num[1005],vis[1005];

int main() {
    int n,k,step = 9999,in;
    scanf("%d%d",&n,&k);
    for(int i = 1;i <= n;i++)
        scanf("%d",&num[i]);

    for(int i = 1;i <= n;i++) {
        int t = i,ans = 0,temp = 1;
        vis[t] = num[t];
        while(t != 1) {
            if(num[t - 1] + k != vis[t]) ans++;
            vis[t - 1] = vis[t] - k;
            if(vis[t - 1] <= 0) {
                temp = 0;
                break;
            }
            t--;
        }

        if(!temp) continue;
        t = i;
        while(t != n) {
            if(vis[t] + k != num[t + 1]) ans++;
            vis[t + 1] = vis[t] + k;
            t++;
        }

        if(ans < step) {
            step = ans;
            in = i;
        }
    }

    printf("%d\n",step);
    int t = in;
    while(t != 1) {
        int ans = num[t - 1] + k - num[t];
        if(ans > 0) printf("- %d %d\n",t - 1,ans);
        if(ans < 0) printf("+ %d %d\n",t - 1,-ans);
        num[t - 1] = num[t] - k;
        t--;
    }

    t = in;
    while(t != n) {
        int ans = num[t] + k - num[t + 1];
        if(ans > 0) printf("+ %d %d\n",t + 1,ans);
        if(ans < 0) printf("- %d %d\n",t + 1,-ans);
        num[t + 1] = num[t] + k;
        t++;
    }

    return 0;
}