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
题意：对一个序列，通过最少次的对某些数据的增或减，使得该序列成为公差为k的等差序列。
思路：遍历所有的数，假设当前数为等差序列中的数，通过增减其它数，将所有的数变为等差序列中的数，记录最小的操作次数，并将操作的数记录下来。注意：调整后序列中的值必须全部大于0，在这跪了好多次。。。
1 #include <stdio.h> 2 #include <string.h> 3 #include <algorithm> 4 #include <iostream> 5 const int N=1005; 6 const int INF=1<<28; 7 using namespace std; 8 int a[N],b[N],f[N]; 9 int main() 10 { 11 int n,d; 12 while(cin>>n>>d){ 13 cin>>f[1]; 14 int flag1 = 1; 15 for (int i = 2; i <= n; i++){ 16 cin>>f[i]; 17 if (f[i]-f[i-1]!=d) 18 flag1 = 0; 19 } 20 if (flag1){ //原序列为公差为d的等差序列 21 puts("0"); 22 continue; 23 } 24 int cnt,Min = INF; 25 for (int i = 1; i <= n; i++){ 26 int m = f[i]-d; 27 cnt = 0;flag1 = 0; 28 for (int j = i-1; j >= 1; j--){//调整f[i]左边的数使其成为公差为d的等差序列 29 if (m <= 0){ //出现小于0的数，则说明该序列不合法 30 flag1 = 1; 31 break; 32 } 33 if (f[j]!=m) 34 cnt++;//记录修改的数的个数 35 m = m-d; 36 } 37 if (flag1) 38 continue; 39 m = f[i]+d; 40 for (int j = i+1; j <= n; j++){//调整f[i]右边的数使其成为公差为d的等差序列 41 if (f[j]!=m) 42 cnt++; 43 m+=d; 44 } 45 if (cnt < Min){ 46 Min = cnt;//记录最少的操作次数 47 m = f[i]-d; 48 memset(a,-1,sizeof(a));//a[]存储对数据进行增的操作 49 memset(b,-1,sizeof(b));//b[]存储对数据进行减的操作 50 for (int j = i-1; j >= 1; j--){ 51 if (f[j] > m) 52 b[j] = f[j]-m; 53 if (f[j] < m) 54 a[j] = m-f[j]; 55 m-=d; 56 } 57 m = f[i]+d; 58 for (int j = i+1; j <= n; j++){ 59 if (f[j] > m) 60 b[j] = f[j]-m; 61 if (f[j] < m) 62 a[j] = m-f[j]; 63 m+=d; 64 } 65 } 66 } 67 cout<<Min<<endl; 68 for (int i = 1; i <= n; i++){ 69 if (a[i]!=-1){ 70 printf("+"); 71 cout<<" "<<i<<" "<<a[i]<<endl; 72 } 73 if (b[i]!=-1){ 74 printf("-"); 75 cout<<" "<<i<<" "<<b[i]<<endl; 76 } 77 } 78 } 79 return 0; 80 }