题目链接:http://codeforces.com/problemset/problem/402/B
题目大意:给出这n棵树的高度,通过增加或是减少树的高度使得第i棵树比第i-1棵树高k米,求最小的步数及每步的操作。
题目分析1:枚举以哪棵树为基准。
代码参考1:
#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<cstdio>
#include<string>
#include<cstring>
#include<sstream>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
const int N = 2000;
int a[N], b[N];
int main()
{
int n, k, i, j;
while (~scanf("%d%d", &n, &k))
{
memset(b, 0, sizeof(b));
for (i = 0; i < n; ++i)
{
scanf("%d", &a[i]);
}
for (i = 0; i < n; ++i)//枚举基准
{
for (j = n - 1; j >= 0; --j)
{
int d = j - i;
if (a[j] - a[i] != d * k)//如果他们的高度差不符合条件,改变次数+1
{
b[i]++;
}
}
if (a[i] - i * k <= 0)//a[0]是最小的数,如果它<=0的话就不符合条件,赋值为INF
{
b[i] = N;
}
}
int m = N, p;
for (i = 0; i < n; ++i)//选出修改次数最小的数和所在位置
{
if (b[i] < m)
{
m = b[i];//最小次数
p = i;//所在位置
}
}
printf("%d\n", m);
for (j = 0; j < n; ++j)//输出每步的具体操作
{
int d = a[j] - a[p] - (j - p - 1) * k;
if (d > k)
{
printf("- %d %d\n", j + 1, d - k);
}
else
if (d < k)
{
printf("+ %d %d\n", j + 1, k - d);
}
}
}
return 0;
}
====================================================================================================================================
代码参考2:(小土豆的~)
#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<cstdio>
#include<string>
#include<cstring>
#include<sstream>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
int cnt[1 << 10];
int arr[1 << 10];
int main()
{
int n, k, i, a;
while (~scanf("%d%d", &n, &k))
{
memset(cnt, 0, sizeof(cnt));
for (i = 1; i <= n; ++i)
{
scanf("%d", &a);
arr[i] = a;
if (a - (i - 1)*k > 0)
{
++cnt[a - (i - 1)*k];//a-(i-1)*k以i为基准1位置该有的高度
//5 1
//1 1 1 2 3
//能让首项为1的有2个,能让首项为2或3的都只有1个,所以首项就是1了
}
}
int ans = 0, base = -1;
for (i = 0; i < 1<<10; ++i)//找出能让首项为i的次数最多的
{
if (cnt[i] > ans)
{
ans = cnt[i];
base = i;
}
}
printf("%d\n", n - ans);
for (i = 1; i <= n; ++i)
{
if (base != arr[i])
{
if (base < arr[i])
{
printf("-");
}
else
{
printf("+");
}
printf(" %d %d\n", i, abs(base - arr[i]));
}
base += k;
}
}
return 0;
}