**E Sleeping Schedule **
題目描述
Vova had a pretty weird sleeping schedule. There are h hours in a day. Vova will sleep exactly n times. The i-th time he will sleep exactly after ai hours from the time he woke up. You can assume that Vova woke up exactly at the beginning of this story (the initial time is 0). Each time Vova sleeps exactly one day (in other words, h hours).
Vova thinks that the i-th sleeping time is good if he starts to sleep between hours l and r inclusive.
Vova can control himself and before the i-th time can choose between two options: go to sleep after ai hours or after ai−1 hours.
Your task is to say the maximum number of good sleeping times Vova can obtain if he acts optimally.
Input
The first line of the input contains four integers n,h,l and r (1≤n≤2000,3≤h≤2000,0≤l≤r<h) — the number of times Vova goes to sleep, the number of hours in a day and the segment of the good sleeping time.
The second line of the input contains n integers a1,a2,…,an (1≤ai<h), where ai is the number of hours after which Vova goes to sleep the i-th time.
Output
Print one integer — the maximum number of good sleeping times Vova can obtain if he acts optimally.
Example Input
7 24 21 23
16 17 14 20 20 11 22
Example Output
3
Note
The maximum number of good times in the example is 3.
The story starts from t=0. Then Vova goes to sleep after a1−1 hours, now the time is 15. This time is not good. Then Vova goes to sleep after a2−1 hours, now the time is 15+16=7. This time is also not good. Then Vova goes to sleep after a3 hours, now the time is 7+14=21. This time is good. Then Vova goes to sleep after a4−1 hours, now the time is 21+19=16. This time is not good. Then Vova goes to sleep after a5 hours, now the time is 16+20=12. This time is not good. Then Vova goes to sleep after a6 hours, now the time is 12+11=23. This time is good. Then Vova goes to sleep after a7 hours, now the time is 23+22=21. This time is also good.
題目大意
一個人每天從0點開始睡覺 ，睡n次覺，每天共h小時，每次可以選擇睡ai時間或者ai - 1時間，當醒來時在l到r區間時，算一次 good time，求maximum number of good sleeping times.
Solution
動態規劃。
每次決策依賴於上一次的時間，定義 dp[ i ][ j ]爲前i次睡覺j次選擇睡ai時間所能獲得的最大good sleeping times。
（或者定義dp[i][j]爲前i次睡覺在j時間醒來的最大good sleeping times）。
按第一種定義方法，先處理一個前i次均按ai睡眠的前綴和。
每次的新時間爲 (num[i] - (i - j) ) % h;
狀態轉移
dp[i][j] = max(dp[i - 1][j] , dp[i - 1][j - 1]) + temp;
第一種是第i次睡覺選擇ai - 1時間。
第二種是第i次睡覺選擇ai時間。
答案即爲dp[n][i]的最大值。
代碼
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long ll;
const int SZ = 2000 + 200;
int num[SZ],x,n,h,l,r,dp[SZ][SZ],ans,a[SZ];
int main()
{
scanf("%d%d%d%d",&n,&h,&l,&r);
for(int i = 1;i <= n;i ++ )
{
scanf("%d",&a[i]);
num[i] = num[i - 1] + a[i];
}
for(int i = 1;i <= n;i ++ )
for(int j = 0;j <= i;j ++ )
{
if(j == 0)
{
int temp = (num[i] - i) % h;
if(temp >= l && temp <= r) temp = 1;
else temp = 0;
dp[i][j] = dp[i - 1][j] + temp;
}
else
{
int temp1;
temp1 = (num[i] - (i - j)) % h;
if(temp1 >= l && temp1 <= r) temp1 = 1;
else temp1 = 0;
dp[i][j] = max(dp[i - 1][j] + temp1,dp[i - 1][j - 1] + temp1);
}
}
for(int i = 0;i <= n;i ++ )
ans = max(ans,dp[n][i]);
printf("%d\n",ans);
return 0;
}
2020.3.14