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>
#include<cstring>
#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<min)
{
min=d;
for(int k=1;k<=n;k++)ans[k]=f[k];
}
}
printf("%d\n",min);
for(int i=1;i<=n;i++)
if(ans[i])
{
if(ans[i]>0)printf("+ %d %d\n",i,ans[i]);
else printf("- %d %d\n",i,ans[i]);
}
}
return 0;
}
Codeforces Round #236 (Div. 2)B,布布扣,bubuko.com
Codeforces Round #236 (Div. 2)B
原文：http://blog.csdn.net/acvcla/article/details/21352925