C - Together

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Time limit?: 2sec /?Memory limit?: 256MB

Score :?300?points

Problem Statement

You are given an integer sequence of length?N,?a1,a2,…,aN.

For each?1≤i≤N, you have three choices: add?1?to?ai, subtract?1?from?ai?or do nothing.

After these operations, you select an integer?X?and count the number of?i?such that?ai=X.

Maximize this count by making optimal choices.

Constraints

• 1≤N≤105
• 0≤ai<105(1≤i≤N)
• ai?is an integer.

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Input

The input is given from Standard Input in the following format:

N a1 a2 .. aN

Output

Print the maximum possible number of?i?such that?ai=X.

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Sample Input 1

Copy 7 3 1 4 1 5 9 2

Sample Output 1

Copy 4

For example, turn the sequence into?2,2,3,2,6,9,2?and select?X=2?to obtain?4, the maximum possible count.

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Sample Input 2

Copy 10 0 1 2 3 4 5 6 7 8 9

Sample Output 2

Copy 3

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Sample Input 3

Copy 1 99999

Sample Output 3

Copy 1

?這個題比較簡單。就是每個數可以進行的操作是加1減一或者不變，貪心取三個就好啊

#include <stdio.h> #include <bits/stdc++.h> using namespace std; const int N=1e5+5; int a[N]; int main() {int n;cin>>n;for(int i=0; i<n; i++){int x;cin>>x;a[x]++;}int ma=a[0]+a[1];for(int i=2;i<N;i++){int t=a[i-2]+a[i-1]+a[i];if(t>ma)ma=t;}cout<<ma<<endl;return 0; }

D - Derangement

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Time limit?: 2sec /?Memory limit?: 256MB

Score :?400?points

Problem Statement

You are given a permutation?p1,p2,…,pN?consisting of?1,2,..,N. You can perform the following operation any number of times (possibly zero):

Operation: Swap two?adjacent?elements in the permutation.

You want to have?pi≠i?for all?1≤i≤N. Find the minimum required number of operations to achieve this.

Constraints

• 2≤N≤105
• p1,p2,..,pN?is a permutation of?1,2,..,N.

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Input

The input is given from Standard Input in the following format:

N p1 p2 .. pN

Output

Print the minimum required number of operations

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Sample Input 1

Copy 5 1 4 3 5 2

Sample Output 1

Copy 2

Swap?1?and?4, then swap?1?and?3.?p?is now?4,3,1,5,2?and satisfies the condition. This is the minimum possible number, so the answer is?2.

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Sample Input 2

Copy 2 1 2

Sample Output 2

Copy 1

Swapping?1?and?2?satisfies the condition.

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Sample Input 3

Copy 2 2 1

Sample Output 3

Copy 0

The condition is already satisfied initially.

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Sample Input 4

Copy 9 1 2 4 9 5 8 7 3 6

Sample Output 4

Copy 3

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繼續簡單題，但是只能交換前后兩個，所以正著掃一遍，倒著掃一遍就好的

#include <stdio.h> #include <bits/stdc++.h> using namespace std; const int N=1e5+5; int a[N]; int main() {int n;cin>>n;int t=0;for(int i=1; i<=n; i++){int x;cin>>a[i];}for(int i=1; i<n; i++){if(i==a[i]){swap(a[i],a[i+1]);t++;}}for(int i=n; i>1; i--){if(i==a[i]){swap(a[i],a[i-1]);t++;}}cout<<t<<endl;return 0; }

E - ConvexScore

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Time limit?: 2sec /?Memory limit?: 256MB

Score :?700?points

Problem Statement

You are given?N?points?(xi,yi)?located on a two-dimensional plane. Consider a subset?S?of the?N?points that forms a convex polygon. Here, we say a set of points?Sforms a convex polygon?when there exists a convex polygon with a positive area that has the same set of vertices as?S. All the interior angles of the polygon must be strictly less than?180°.

For example, in the figure above, {A,C,E} and {B,D,E} form convex polygons; {A,C,D,E}, {A,B,C,E}, {A,B,C}, {D,E} and {} do not.

For a given set?S, let?n?be the number of the points among the?N?points that are inside the convex hull of?S?(including the boundary and vertices). Then, we will define the?score?of?S?as?2n?|S|.

Compute the scores of all possible sets?S?that form convex polygons, and find the sum of all those scores.

However, since the sum can be extremely large, print the sum modulo?998244353.

Constraints

• 1≤N≤200
• 0≤xi,yi<104(1≤i≤N)
• If?i≠j,?xi≠xj?or?yi≠yj.
• xi?and?yi?are integers.

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Input

The input is given from Standard Input in the following format:

N x1 y1 x2 y2 : xN yN

Output

Print the sum of all the scores modulo?998244353.

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Sample Input 1

Copy 4 0 0 0 1 1 0 1 1

Sample Output 1

Copy 5

We have five possible sets as?S, four sets that form triangles and one set that forms a square. Each of them has a score of?20=1, so the answer is?5.

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Sample Input 2

Copy 5 0 0 0 1 0 2 0 3 1 1

Sample Output 2

Copy 11

We have three "triangles" with a score of?1?each, two "triangles" with a score of?2?each, and one "triangle" with a score of?4. Thus, the answer is?11.

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Sample Input 3

Copy 1 3141 2718

Sample Output 3

Copy 0

There are no possible set as?S, so the answer is?0.

給出一些點，每個凸多邊形的貢獻為2^n?|S|，求貢獻和。

暴力枚舉判斷就好了

#include<cstdio> int n,ans,x[210],y[210],f[210]; const int MD=998244353; int main() {scanf("%d",&n);f[0]=1;for(int i=1; i<=n; i++){scanf("%d%d",&x[i],&y[i]);f[i]=1ll*f[i-1]*2%MD;}ans=f[n]-n-1;for(int i=1; i<=n; i++){for(int j=i+1; j<=n; j++){int d=0;for(int k=j+1; k<=n; k++)d+=(y[i]-y[j])*(x[i]-x[k])==(y[i]-y[k])*(x[i]-x[j]);ans=(ans+MD-f[d])%MD;}}printf("%d\n",ans);return 0; }

?

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轉載于:https://www.cnblogs.com/BobHuang/p/7471035.html

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