Drawing Book

水题，推公式

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <string>
#include <bitset>
#include <cstdio>
#include <limits>
#include <vector>
#include <climits>
#include <cstring>
#include <cstdlib>
#include <fstream>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <unordered_map>

using namespace std;


int main(){
    int n;
    cin >> n;
    int p;
    cin >> p;
    cout<<min(p/2,(n-p)/2)<<endl;
    return 0;
}

Tailor Shop

有n个群，每个群有规定的最小的钱，每个按钮有规定的价格，问最少的使用的按钮的数量

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;
const int maxn = 1e5+100;
int a[maxn];
int n,p;
int main() {
    scanf("%d %d",&n,&p);
    for(int i = 0 ;i<n;i++) scanf("%d",&a[i]);
    sort(a,a+n);
    LL num = 1;
    LL ans = 0;
    for(int i = 0;i<n;i++) {
        while(num*p<a[i])  num++;
        ans+=num;
        num++;
    }
    printf("%lld\n",ans);
    return 0;
}

Hackonacci Matrix Rotations

定义$hackonacci(n) = 1\times hackonacci(n-1)+2 \times hackonacci(n-2)+3\times hackonacci(n-3)$并且$hackonacci(1)=1,hackonacci(2)=2,hackonacci(3)=3$,对于矩阵中的位置的价值为$hackonacci((i\times j)^2)$如果是奇数为Y,否则为X,通过打表我们会发现一些奇特的规律，然后就会发现这道的做法

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;

int n,q;
int ln[2001][2001];
int a[2001][2001];
int ans[5],Angle;
int b[] = {0,1,0,1,0,0,1,1};
int Get(int i,int j) {
    int ant = (i*j)%7;
    ant = (ant*ant)%7;
    return ant == 0?7:ant;
}
void Ratote() {
    for(int i = 1;i<=(n/2);i++) {
        for(int j = 1;j<=n;j++)
            swap(a[i][j],a[n+1-i][j]);
    }
    for(int i = 1;i<=n;i++) {
        for(int j = 1;j<=i;j++) swap(a[i][j],a[j][i]);
    }
}
int Diff() {
    int ant = 0;
    for(int i = 1;i<=n;i++)
        for(int j = 1;j<=n;j++)
            ant+=(ln[i][j] == a[i][j]?0:1);
    return ant;
}
int main() {
    scanf("%d %d",&n,&q);
    for(int i = 1;i<=n;i++) {
        for(int j = 1;j<=n;j++) {
            a[i][j] = b[Get(i,j)];
            ln[i][j] = a[i][j];
        }
    }
    ans[0] = 0;
    for(int i =1;i<=3;i++) {
        Ratote();
        ans[i] = Diff();
    }
    while(q--) {
        scanf("%d",&Angle);
        (Angle/=90)%=4;
        printf("%d\n",ans[Angle]);
    }
    return 0;
}

Zero-Move Nim

改变的Nim博弈，对于非空的石堆，有一次不用拿的机会。手写几个SG就会发现做法的

#include <bits/stdc++.h>

using namespace std;

int T,n;

int main() {
    scanf("%d",&T);
    while(T--) {
        scanf("%d",&n);
        int ans = 0,data;
        for(int i = 1;i<=n;i++) {
            scanf("%d",&data);
            ans ^= data+(data%2==0?-1:1);
        }
        if(ans) printf("W\n");
        else printf("L\n");

    }
    return 0;
}