題目：

A. Mr. Kitayuta, the Treasure Hunter time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

The Shuseki Islands are an archipelago of 30001 small islands in the Yutampo Sea. The islands are evenly spaced along a line, numbered from 0 to 30000 from the west to the east. These islands are known to contain many treasures. There are n gems in the Shuseki Islands in total, and the i-th gem is located on island p_i.

Mr. Kitayuta has just arrived at island 0. With his great jumping ability, he will repeatedly perform jumps between islands to the east according to the following process:

• First, he will jump from island 0 to island d.
• After that, he will continue jumping according to the following rule. Let l be the length of the previous jump, that is, if his previous jump was from island prev to island cur, let l?=?cur?-?prev. He will perform a jump of length l?-?1, l or l?+?1 to the east. That is, he will jump to island (cur?+?l?-?1), (cur?+?l) or (cur?+?l?+?1) (if they exist). The length of a jump must be positive, that is, he cannot perform a jump of length 0 when l?=?1. If there is no valid destination, he will stop jumping.

Mr. Kitayuta will collect the gems on the islands visited during the process. Find the maximum number of gems that he can collect.

Input

The first line of the input contains two space-separated integers n and d (1?≤?n,?d?≤?30000), denoting the number of the gems in the Shuseki Islands and the length of the Mr. Kitayuta's first jump, respectively.

The next n lines describe the location of the gems. The i-th of them (1?≤?i?≤?n) contains a integer p_i (d?≤?p₁?≤?p₂?≤?...?≤?p_n?≤?30000), denoting the number of the island that contains the i-th gem.

Output

Print the maximum number of gems that Mr. Kitayuta can collect.

Examples Input Copy 4 10
10
21
27
27 Output Copy 3 Input Copy 8 8
9
19
28
36
45
55
66
78 Output Copy 6 Input Copy 13 7
8
8
9
16
17
17
18
21
23
24
24
26
30 Output Copy 4 Note

In the first sample, the optimal route is 0 ?→? 10 (+1 gem) ?→? 19 ?→? 27 (+2 gems) ?→?...

In the second sample, the optimal route is 0 ?→? 8 ?→? 15 ?→? 21?→? 28 (+1 gem) ?→? 36 (+1 gem) ?→? 45 (+1 gem) ?→? 55 (+1 gem) ?→? 66 (+1 gem) ?→? 78 (+1 gem) ?→?...

In the third sample, the optimal route is 0 ?→? 7 ?→? 13 ?→? 18 (+1 gem) ?→? 24 (+2 gems) ?→? 30 (+1 gem) ?→?...

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思路和實現都不難的動態規劃

(._. )

做的時候沒看出來長度的限制 擔心復雜度太大所以不敢寫

(._. )

菜雞

(._. )

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dp[i][j]表示到編號為 i 的島且上次跳躍長度為 j 時能取到的最多的gems的數目。

注意：島的編號限制在30000以內，且每次最多增長一步，第一步跳躍長度為d，總的跳躍長度 = d?+?(d?+?1)?+?(d?+?2)?+?...?+?(d?+?245)? ≥? 1?+?2?+?...?+?245?=?245·(245?+?1)?/?2 ?=? 30135 ?>? 30000。所以跳躍長度最長為（d+245），最短為（d-245），因此 j 的枚舉長度在[d-245，d+245]之間，第二維的空間縮小到500。

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#include<iostream> #include<cstring> #include<cstdio> #include<algorithm> #include<set>using namespace std;const int maxn = 3*1e4+10;int dp[maxn][510];; int num[maxn];int main() {int base;int a, b;int n, m, d;int i, j, k;scanf("%d %d",&n,&d);for(i = 1; i <= n; i++)scanf("%d", &a), num[a]++;base = max(d -250,0);memset(dp, -1, sizeof dp);dp[d][d-base] = num[d];int ans=0;for(i = d ; i <= 30000; i++){for( j = 1 ;j <= 500; j++){if(dp[i][j] == -1) continue;for(k = -1; k < 2; k++){if( j + k < 1|| base + i + j + k > 30000) continue;dp[i + base + j + k][j + k] = max(dp[base + i + j + k][j + k], dp[i][j] + num[i + j + k + base]);}ans = max(ans, dp[i][j]);}}cout << ans << endl;return 0; } View Code

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轉載于:https://www.cnblogs.com/orange-/p/10791335.html

超強干貨來襲 云風專訪：近40年碼齡，通宵達旦的技術人生

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