zstu新生赛

来源：互联网发布：什么叫大数据思维编辑：程序博客网时间：2024/05/16 05:29

4238: Save the Princess

博弈题，直接判断n奇偶以及k的位置是否是边界。

#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: 极差

#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: 牛吃草

这道题奇怪的被卡，由于比赛时没有直接出结果，没发现错误。结束了拍了一下，发现前半部分写的有问题。。。。

#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: 众数

#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的个数，每次去掉最大的，看最后结果情况。

#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，比赛时没有出评测结果，结果就GG了，赛后写的。
等待补充。。。

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