ZOJ 3741 Eternal Reality [dp]

In Academy City, most students have special abilities. Such as Railgun, Teleport, Telekinesis, AIM Stalker and so on. Here, AIM (An Involuntary Movement) is a term used to refer to the phenomenon in which an esper involuntarily produces an invisible energy field around the esper. Different students have different type of AIM dispersion field, so they also have different level of abilities.

Of course, a higher level students can often deal with more issues than a lower level students. To classify the students in Academy City, there are 7 levels in total:

Level	Term	Description
Level 0	Person with No Powers	Most students of this level can't keep up at school. They might possess some degree of power, but unable to truly control it.
Level 1	Person with Low Powers	Powers of the degree to bend a spoon, many students belong here.
Level 2	Person with Unusual Powers	Just like Level 1, powers are not very useful in everyday life.
Level 3	Person with Strong Powers	The degree when powers are considered convenient in everyday life, ability-wise this is the Level when one starts to be treated as part of the elite.
Level 4	Person with Great Powers	Powers of an extent that their owner acquires tactical value of a military force.
Level 5	Person with Super Powers	Powers of an extent that their owner can fight alone against a military force on equal terms.
Level 6	Person with Absolute Powers	Powers of an extent that they're considered immeasurable. However, no one can achieve this level, even with the help of Level Upper (it has no effect on persons with super powers). Since this, many institutions have been doing long-term researches about it, such as the Radio Noise Project.

You are a student of level L in Academy City and you are going to take part in a sports competition. The competition consists of N consecutive matches. If you want to get a point in the i-th match, your must reach at least A_i level. According to the rules, you must compete in all matches one by one.

To tell the truth, it won't be easy to compete with so many high-level opponents. Fortunately, you got a special item called Level Upper. Generally, it can increase your level by 1 for a short time. If you use the Level Upper before the i-th match, it's effect will last during the matches [i, i + X - 1]. But it also has a side effect that will make your level become 0 during the matches [i + X, i + X + Y - 1]. After the side effect ends, your level will return to L and you can use the Level Upper again.

Please calculate the maximal points you can get if you properly use the Level Upper.

Input

There are multiple test cases (plenty of small cases with several large cases). For each test case:

The first line contains four integers L (0 <= L <= 5), N, X and Y (1 <= N, X, Y <= 100). The next line contains N integers A_i (0 <= A_i <= 6).

Output

For each test case, output the maximal points you can get.

Sample Input

3 6 1 2
1 3 4 5 6 4

Sample Output

4

Hint

Read the problem description carefully.

题意：有n个关卡，每关一个难度ai，一个等级为L的人要连续闯关，当L>=ai时他就可以得1分，否则不得分（继续下一关）。喝一次药可以使他的等级+1，持续x关，之后会出现副作用，他的等级变成0，持续y关，再过后又回复正常，可以继续喝药。问他最多得多少分。关卡难度有0～6，人的等级L为0~5。

注意一点，就算喝药也不能到达6级！！

d[i][j]表示前i关，最后一次在第j关喝药可以得到的最高分。

d[i][j] = d[i-1][j] + 第i关得分（视药效情况）      , j < i

d[i][i] = max{ d[i-1][k] } + (L+1 >= a[i])     , k <= i - x - y

一开始判断如果L等于5那就直接统计难度小于等于5的关卡数就行。

#include<cassert>
#include<algorithm>
#include<cmath>
#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<set>
#include<queue>
#include<map>
using namespace std;
#define rep(i,f,t) for(int i = (f), _end = (t); i <= _end; ++i)
#define clr(c,x) memset(c,x,sizeof(c));
#define debug(x) cout<<"debug  "<<x<<endl;
const int INF = 0x3f3f3f3f;
typedef long long int64;
//*******************************************************************************

const int maxn = 110;
int a[maxn];
int d[maxn][maxn];


int main(){
    //freopen("3741.in","r",stdin);
    int lv,n,x,y;
    while(~scanf("%d%d%d%d",&lv,&n,&x,&y)){
        rep(i,1,n) scanf("%d",a+i); 
        if(lv == 5){
            int ans = 0;
            rep(i,1,n)if(a[i] <= 5)++ans;
            printf("%d
",ans);
            continue;
        }

        d[1][0] = (lv >= a[1]);
        d[1][1] = (lv+1 >= a[1]);
        rep(i,2,n){
            d[i][0] = d[i-1][0] + (lv >= a[i]);   //d[i][0]表示从没喝药
            rep(j,1,i-1){
                d[i][j] = d[i-1][j];
                if(i < j+x)
                    d[i][j] += (lv+1 >= a[i]);        //有效期
                else if(i < j+x+y)
                    d[i][j] += (0 >= a[i]);          //副作用期
                else
                    d[i][j] += (lv >= a[i]);          //已经回复正常
            }
            d[i][i] = d[i-1][0] + (lv+1 >= a[i]);
            rep(k,1,i-x-y){
                d[i][i] = max( d[i][i], d[i-1][k] + (lv+1 >= a[i]) );
            }
        }

        int ans = 0;
        rep(j,0,n){
            ans = max(ans, d[n][j]);
        }
        printf("%d
",ans);
    }
    return 0;
}

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