6576: guruguru
时间限制: 1 Sec 内存限制: 128 MB
提交: 44 解决: 19
[提交] [状态] [讨论版] [命题人:admin]
题目描述
Snuke is buying a lamp. The light of the lamp can be adjusted to m levels of brightness, represented by integers from 1 through m, by the two buttons on the remote control.
The first button is a "forward" button. When this button is pressed, the brightness level is increased by 1, except when the brightness level is m, in which case the brightness level becomes 1.
The second button is a "favorite" button. When this button is pressed, the brightness level becomes the favorite brightness level x, which is set when the lamp is purchased.
Snuke is thinking of setting the favorite brightness level x so that he can efficiently adjust the brightness. He is planning to change the brightness n−1 times. In the i-th change, the brightness level is changed from ai to ai+1. The initial brightness level is a1. Find the number of times Snuke needs to press the buttons when x is set to minimize this number.
Constraints
2≤n,m≤105
1≤ai≤m
ai≠ai+1
n, m and ai are integers.
输入
Input is given from Standard Input in the following format:
n m
a1 a2 … an
输出
Print the minimum number of times Snuke needs to press the buttons.
样例输入
4 6 1 5 1 4
样例输出
5
题意：两个按钮, 一个可以使计数器+1(计数器数字从1-m), 当前值为m时, 再+1就变成了1
另一个按钮储存了一个值x, 按一下就从任意值会变成x
n-1次操作, 由一个数组a[n]描述, 第i次操作: 将计数器从a[i]调到a[i+1]
将x设置为某个值, 使得所有操作需要按按钮的次数总和最小, 输出这个最小值
思考： 对于L=a[i] 和 R=a[i+1]
考虑 ①：R==L 不需要按 没啥用 ② ： R-L==1 无论X是什么也没有用 ③ R-L>1 那么对于存在一个X在L-R区间
假如X=L+2可以减少按键一次 ~~~~~~~~~~~~~ X=R 可以减少R-L-1次，对于L>R的情况考虑R=R+m.这样就不需要考虑越界问题
因此维护X从1到m的值即可。ans记录如果都是按一下需要的最大总数。Ans记录最大的如果X在某位置的需要减少的次数
Ans=max(Ans,sum[i]+sum[i+m])(1<=i<=m) 因为有R=R+m 所以是i+m。这样的话暴力复杂度为n*m
所以利用前缀和优化成O(1)
我们可以这样: p[l+2] += 1, p[r+1] -= (r-(l+2)+1)+ 1, p[r+2] += r-(l+2)+1, 每次只更新这三个值, 最后再从头到尾p[i] += p[i-1]
具体的一组例子, 比如l=0, r=5(emmm, 假设有0这个亮度，题意借鉴而来就不修改了)
| 更新操作 | 复杂度 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|---|---|
| 更新三个值 | O(1)O(1) | 0 | 1 | 0 | 0 | 0 | -5 | 4 |
| p[i]+=p[i-1] | O(n)O(n) | 0 | 1 | 1 | 1 | 1 | -4 | 0 |
| p[i]+=p[i-1] | O(n)O(n) | 0 | 1 | 2 | 3 | 4 | 0 | 0 |
时间复杂度(n+m)
//Sinhaeng Hhjian
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<vector>
#include<stdlib.h>
#include<math.h>
#include<set>
#include<algorithm>
using namespace std;
typedef long long ll;
const int mod = 998244353;
const int maxn = 1e5+5;
ll sum[maxn*2];
int main()
{
int n, m, temp, x;
scanf("%d%d%d", &n, &m, &temp);
ll ans=0;
for(int i=1;i<n;i++){
scanf("%d", &x);
ll l=temp, r=x;
if(l>r)
r+=m;
ans+=r-l;
if(r-l>1){
sum[l+2]+=1;
sum[r+1]-=(r-(l+2)+1)+1;
sum[r+2]+=r-(l+2)+1;
}
temp=x;
}
for(int i=1;i<=2*m;i++)
sum[i]+=sum[i-1];
for(int i=1;i<=2*m;i++)
sum[i]+=sum[i-1];
ll ans1=0;
for(int i=1;i<=m;i++)
ans1=max(ans1, sum[i]+sum[i+m]);
printf("%lld\n", ans-ans1);
return 0;
}