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西安理工大学2024年程序设计校赛（校外同步赛）

A

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#include<bits/stdc++.h>
using namespace std;
 
 
signed main()
{
    char c;
    cin >> c;
    if(c == 'A' || c == 'B' || c == 'C')
        cout << "YES" << endl;
    else cout << "NO" << endl;
    return 0;
}

B

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牛马（）

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#include<bits/stdc++.h>
using namespace std;
 
 
signed main()
{
    string a = "", b = "", c = "";
    string s;
    getline(cin,a);
    getline(cin,b);
    getline(cin,c);
    if(a == "Vertebrates"){
        if(b == "Bird"){
            if(c == "Haunted Den"){
                cout << "Pigeons" << endl;
            }
            else{
                cout << "Eagle" << endl;
            }
        }
        else{
            if(c == "Haunted Den"){
                cout << "People" << endl;
            }
            else{
                cout << "Cow" << endl;
            }
        }
    }
    else{
        if(b == "Insect"){
            if(c == "Herbivorous"){
                cout << "Caterpillar" << endl;
            }
            else{
                cout << "Fleas" << endl;
            }
        }
        else{
            if(c == "Haunted Den"){
                cout << "Worms" << endl;
            }
            else{
                cout << "Leeches" << endl;
            }
        }
    }
    return 0;
}

C

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结论题目（我才不写高精度）

给左边式子*2，然后均值不等式

左边大于或等于右边

等于当且仅当三者相等

image-20240320200757189 
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#include <bits/stdc++.h>
 
using namespace std;
 
int main(){
    string x,y,z; cin>>x>>y>>z;
    if(x==z&&x==y) cout<<"=";
    else cout<<">";
    return 0;
}

D

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后手直接走对角线

然后直接算最多能走多少个对角线

然后如果还能继续走

那就后手输了

否则后手赢了

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#include<bits/stdc++.h>
using namespace std;
 
int main()
{
    long long int t,d,k,i,ans,p;
    cin >> t;
    for(i=0;i<t;i++)
    {
        cin >> d >> k;
         
        ans=d/(k*sqrt(2));
            long double x=ans*k,y=x;
             
            if((x+k)*(x+k)+y*y<=d*d)
            {
                cout<<"Parry"<<endl;
            }
            else
                cout<<"Mercedes"<<endl;
    }
     
    return 0;
}

E

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大概看看就行（反正也是粘贴的）掌握一下大概

暴力枚举3个点

然后计算面积（如果面积为0就退出，然后计算圆的半径）

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#include "bits/stdc++.h"

double solve(int x0, int y0, int x1, int y1, int x2, int y2) {
    x1 -= x0;
    y1 -= y0;
    x2 -= x0;
    y2 -= y0;
    int S = std::abs(x1 * y2 - x2 * y1);
    if (S == 0)
        return 0.;
        
    double a[3] = {std::sqrt(x1 * x1 + y1 * y1), std::sqrt(x2 * x2 + y2 * y2),
                    std::sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2))};
    std::sort(a, a + 3);
    double cos1 = (a[0] * a[0] + a[2] * a[2] - a[1] * a[1]) / (2 * a[0] * a[2]);
    double cos2 = (a[1] * a[1] + a[2] * a[2] - a[0] * a[0]) / (2 * a[1] * a[2]);
    double fenmu = std::sqrt((1 + cos1) / (1 - cos1)) + std::sqrt((1 + cos2) / (1 - cos2)) + 2;
    double ans = a[2] / fenmu;
    if (std::isinf(ans))
        return 0.;
    return ans;
}

int main() {
    std::ios::sync_with_stdio(0);
    std::cin.tie(0);

    int n;
    std::cin >> n;
    std::vector<std::pair<int,int> > a(n);
    for (int i = 0; i < n; i ++) {
        std::cin >> a[i].first >> a[i].second;
    }
    double ans = 0.;
    for (int i = 0; i < n; i ++) {
        for (int j = i + 1; j < n; j ++) {
            for (int k = j + 1; k < n; k ++) {
                ans = std::max(ans, solve(a[i].first, a[i].second, a[j].first, a[j].second, a[k].first, a[k].second) );
            }
        }
    }
    
    std::cout << std::fixed << std::setprecision(6);
    std::cout << ans << "\n";
    
    return 0;
}

F

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树状数组太适合了阿喂

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#include <bits/stdc++.h>
using namespace std;

int sum[1000005];
int lowbit(int x) {
	return x & -x;
}
void add(int x, int k) {
	while (x <= 1000000) {
		sum[x] += k;
		x += lowbit(x);
	}
}
int query(int x) {
	int res = 0;
	while (x) {
		res += sum[x];
		x -= lowbit(x);
	}
	return res;
}
//树状数组操作
int a[200005];

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	int n, q, m;
	cin >> n >> q >> m;
	long long ans = 0;
	for (int i = 1 ; i <= n ; i++) {
		cin >> a[i];
		add(a[i], 1);
		ans += query(m / a[i]);
        //先计算出来
	}
	while (q--) {
		int op;
		cin >> op;
		if (op == 1) {
			int pos, k;
			cin >> pos >> k;
			ans -= query(m / a[pos]);
            //减去之前的
			add(a[pos], -1);
            //修改掉
			a[pos] = k;
            //改掉
			add(a[pos], 1);
            //上传新的
			ans += query(m / a[pos]);
            //重新计入答案
		} else {
			cout << ans << "\n";
		}
	}
	return 0;
}

G

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直接dfs

注意是从高位向低位dfs

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#include<iostream>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
 
long long m,n;
void dfs(long long x)
{
    int y = x % 10;
    if(x > 10) m++;
    if(y - 2 >= 0 && (x * 10 + y - 2) <= n) dfs(x * 10 + y - 2);
 
    if(y + 2 <= 9 && (x * 10 + y + 2) <= n) dfs(x * 10 + y + 2);
}
 
void solve()
{
    cin >> n;
    for(int i = 1;i < 10;i++) dfs(i);
    cout << m << '\n';
}
 
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t = 1;
    //cin >> t;
    while(t--) solve();
    return 0;
}

I

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显然确实如此（）

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#include<bits/stdc++.h>
using namespace std;
#define ll long long
int main(){
    int t;
    cin>>t;
    while(t--){
        int n;
        cin>>n;
        
        ll sum=0,flag=0;
        for(int i=0;i<n;i++){
            ll x;
            
         cin>>x;
            if(x%2==1)flag=1;
        }
        if(flag){
            cout<<"halo"<<endl;
        }
        else {
            cout<<"parry"<<endl;
        }
    }
    return 0;
}

H

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直接二分那个值

然后算出最大的需要增加的

看看会不会增加到0就可以了

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#include<bits/stdc++.h>
using namespace std;
int n,x[100005],y[100005];
bool check(int mid)
{
    int l=1e9,r=-1e9,maxx=0;
    for(int i=1;i<=n;i++)
    {
        if(y[i]<mid)
        {
            maxx=max(maxx,mid-y[i]);
            l=min(l,x[i]);
            r=max(r,x[i]);
        }
    }
    for(int i=1;i<=n;i++)
    {
        if(x[i]>=l&&x[i]<=r)
        {
            if(y[i]+maxx>0)
            {
                return false;
            }
        }
    }
    return true;
}
signed main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        cin>>x[i]>>y[i];
    }
    int l=-1e9,r=0;
    while(l<r)
    {
        int mid=l+r+1>>1;
        if(check(mid))
        {
            l=mid;
        }
        else
        {
            r=mid-1;
        }
    }
    cout<<l;
    return 0;
}

L

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#include <bits/stdc++.h>
using namespace std;
vector<char>ch;
int main()
{
    ios::sync_with_stdio(0);
    int n,m,cnt = 1;
    cin>> n>>m;
    string s;
    cin >> s;
    while(m--)
    {
        int op;cin >>op;
        if(op == 1) cnt++;
        else if(op == 2){
            int p ;
            char c;
            cin >> p >> c;
            if((p == 1 && cnt %2 ==1) || (p==2 && cnt % 2 ==0) ) s.insert(s.begin(),c);
            else if((p == 2 && cnt %2 ==1) || (p == 1 && cnt %2 ==0)) s.push_back(c);
        }
    }
    if(cnt %2==0){
        for(int i = s.size()-1;i>=0;i--)cout << s[i];
    }
    else cout << s;
    return 0;
}
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