C	Draw Grids

 

博弈论

 

可以发现，两人最后总是可以走n*m-1步，判断n*m-1的奇偶性即可，且样例比y较小，打表也可~

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int n,m;
 4 int main()
 5 {
 6     cin>>n>>m;
 7     n*=m;
 8     n--;
 9     if(n%2)puts("YES");
10     else
11         puts("NO");
12     return 0;
13 }

 

D	Er Ba Game

 

签到题

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int T;
 4 int a1,a2,b1,b2;
 5 int main()
 6 {
 7     cin>>T;
 8     while(T--)
 9     {
10         cin>>a1>>b1>>a2>>b2;
11         if(a1>b1)swap(a1,b1);
12         if(a2>b2)swap(a2,b2);
13         if(a1==a2&&b1==b2)puts("tie");
14         else if(a1==2&&b1==8)puts("first");
15         else if(a2==2&&b2==8)puts("second");
16         else if(a1==b1&&a2==b2)puts(a1>a2?"first":"second");
17         else if(a1==b1)puts("first");
18         else if(a2==b2)puts("second");
19         else if(((a1+b1)%10)!=((a2+b2)%10))puts(((a1+b1)%10)>((a2+b2)%10)?"first":"second");
20         else if(b1!=b2)puts(b1>b2?"first":"second");
21         else
22             puts("tie");
23     }
24     return 0;
25 }

 

 

F	Girlfriend

 

高中数学

求两个球的交集体积，有模板：https://blog.csdn.net/luyehao1/article/details/86583384

#include <bits/stdc++.h>

typedef long long ll;

const long double pi = acos(-1);

typedef struct {
    long double x, y, z, r;
}Point;

int n;
Point other;
Point s;

//两点之间距离
long double dis(Point p, Point q) {
    long double ans = sqrt((p.x - q.x)*(p.x - q.x) + (p.y - q.y)*(p.y - q.y) + (p.z - q.z)*(p.z - q.z));
    return ans;
}
long double x1,x2,yy1,y2,z1,z2,r1,r2,a[2],b[2],c[2],d[2],e[2],f[2];
long double k1,k2;
long double pw(long double x){
    return x*x;
}
long double calr(long double a,long double b, long double k){
    return (-a*a+k*k*b*b)/(1-k*k) + pw((-k*k*b+a)/(1-pw(k)));
}
 long double calpos( long double a,long double b, long double k){
    return (a-pw(k)*b)/(1-pw(k));
}
int main()
{
    int T;
    scanf("%d", &T);
    while (T--)
    {
        for(int i=0;i<2;i++)
            scanf("%Lf%Lf%Lf%Lf%Lf%Lf",&a[i],&b[i],&c[i],&d[i],&e[i],&f[i]);
        scanf("%Lf%Lf",&k1,&k2);
        s.x=calpos(a[0],d[0],k1);
        s.y=calpos(b[0],e[0],k1);
        s.z=calpos(c[0],f[0],k1);
        long double r2 = calr(a[0],d[0],k1)+calr(b[0],e[0],k1)+calr(c[0],f[0],k1);
        s.r=sqrt(r2);
        other.x=calpos(a[1],d[1],k2);
        other.y=calpos(b[1],e[1],k2);
        other.z=calpos(c[1],f[1],k2);
        r2=calr(a[1],d[1],k2)+calr(b[1],e[1],k2)+calr(c[1],f[1],k2);
        other.r=sqrt(r2);
        long double ans = 0;
        long double d = dis(s, other);
        if(other.r>s.r) std::swap(other,s);
        if (d >= s.r + other.r) {

        }else if (d + other.r <= s.r) {
            ans += (4.0 / 3)*pi*other.r*other.r*other.r;
        }
        else {
            long double co = (s.r*s.r + d * d - other.r*other.r) / (2.0*d*s.r);
            long double h = s.r*(1 - co);
            ans += (1.0 / 3)*pi*(3.0*s.r - h)*h*h;
            co = (other.r*other.r + d * d - s.r*s.r) / (2.0*d*other.r);
            h = other.r*(1 - co);
            ans += (1.0 / 3)*pi*(3.0*other.r - h)*h*h;
        }
        printf("%.3Lf\n",ans);
    }
    return 0;
}

 

I

Penguins

 

经典bfs

 有点长，先放着~

 

K

Stack

 

****放着，大佬写的，还是看不懂，先把蓝书看完~