zstu新生賽

4238: Save the Princess

博弈題，直接判斷n奇偶以及k的位置是否是邊界。

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int main(){
    int T,n,k;
    cin>>T;
    while(T--){
        cin>>n>>k;
        if(k!=1&&k!=n+1) puts(n%2?"BH":"LYF");
        if(k==1||k==n+1) puts("LYF");
    }
    return 0;
}

4240: 極差

#include<bits/stdc++.h>
using namespace std;
int main(){
    int T;
    cin>>T;
    while(T--){
        int n,x;
        scanf("%d",&n);
        int mx=-1,mn=INT_MAX;
        cout<<mn<<endl;
        for(int i=1;i<=n;i++){
            scanf("%d",&x);
            if(mx<x) mx=x;
            if(mn>x) mn=x;
        }
        printf("%d\n",mx-mn);
    }

    return 0;
} 

4243: 牛喫草

這道題奇怪的被卡，由於比賽時沒有直接出結果，沒發現錯誤。結束了拍了一下，發現前半部分寫的有問題。。。。

#include <cstdio>
#include <iostream>
#include <cmath>
using namespace std;
const double pi = acos(-1.0);
const double eps = 1e-8;
double get(long long x0,long long y0,long long x1,long long y1) {
    return (double)sqrt( (x1-x0)*(x1-x0)*1.0 + (y1-y0)*(y1- y0)*1.0 );
}
int dsgn(double x){return x < -eps ? -1 : x > eps;}
double area(long long x0,long long y0,double r1,long long x1,long long y1,double r2){
    double d =  get(x0,y0,x1,y1);
    double a1=acos((r1*r1+d*d-r2*r2)/(2.0*r1*d));
    double a2=acos((r2*r2+d*d-r1*r1)/(2.0*r2*d));
    return (a1*r1*r1+a2*r2*r2-r1*d*sin(a1));
}
double Area(double r, double R, double l){
    if(dsgn(l - r - R) >= 0) return 0;
    else if(dsgn(l - fabs(r - R)) <= 0){
        if(r > R) r = R;
        return pi * r * r;
    }
    double a = acos((l * l + r * r - R * R) / (2 * l * r));
    double b = acos((l * l + R * R - r * r) / (2 * l * R));
    double s1 = a * r * r, s2 = b * R * R;
    double S1 = r * r * sin(a) * cos(a), S2 = R * R * sin(b) * cos(b);
    return s1 + s2 - S1 - S2;
}
int main(){
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    long long T;
    cin>>T;
    while(T--){
        long long x0,y0,x1,y1,r;
        cin>>x0>>y0>>x1>>y1>>r;
        double s1 = pi * r * r;
        double dist = get(x0,y0,x1,y1);
        double s2 = s1 / 2.0;
        double r_=sqrt(s2/pi);
        double lr_=max(0.0,dist-r),rr_=1e5,ansr;
        while(dsgn(rr_-lr_)>0){
            double midr_=(lr_+rr_)/2.0;
            double s_=area(x0,y0,r,x1,y1,midr_);
            if(dsgn(2.0 * Area(r, midr_, dist) - pi * r * r) < 0){
                lr_ = midr_+0.000000001;
                ansr=midr_;
            }
            else rr_=midr_-0.000000001;
        }
        printf("%.4lf\n",ansr);
    }
    return 0;
}

4244: 衆數

#include <bits/stdc++.h>
using namespace std;
int stu[1111];
int ans[1111];
int main(){
    int T;
    cin>>T;
    while(T--){
        memset(stu,0,sizeof(stu));
        int n,x;
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
            scanf("%d",&x);
            stu[x]++; 
        }
        int mx=0;
        ans[0]=0;
        for(int i=0;i<=1000;i++){
            if(mx<stu[i]){
                mx=stu[i];
                ans[0]=1;
                ans[ans[0]]=i;
            }
            else if(mx==stu[i]){
                ans[0]++;
                ans[ans[0]]=i;
            }
        }
        for(int i=1;i<=ans[0];i++){
            if(i-1) printf(" ");
            printf("%d",ans[i]);
        }
        puts("");
    }

    return 0;
} 

4245: KI的斐波那契

斐波那契，f[n-1]和f[n]表示的恰好是a+b和a的個數，每次去掉最大的，看最後結果情況。

#include <bits/stdc++.h>
#define LL long long 
using namespace std;
LL read(){  
    LL x=0; char ch=getchar();  
    while (ch<'0' || ch>'9') ch=getchar();  
    while (ch>='0' && ch<='9'){ x=x*10+ch-'0'; ch=getchar(); }  
    return x; 
}
LL f[99];
bool dfs(LL n,LL m){
    if(n==2) return m==1;
    if(n==1) return 1;
    if(m> f[n-1]) dfs(n-2,m-f[n-1]);
    else dfs(n-1,m);
}
int main(){
    LL T,n,m;
    cin>>T;
    f[0]=1;f[1]=1;
    for(int i=2;i<=91;i++) f[i]=f[i-1]+f[i-2];
    while(T--){
        n=read();m=read();
        puts(dfs(n,m)?"a":"b");
    }
    return 0;
}

吐槽一下，比賽時沒有出評測結果，結果就GG了，賽後寫的。
等待補充。。。