NOIP 模拟 $25\; \rm queen$

题解 \(by\;zj\varphi\)

这是一道纯分类讨论，然后推式子的题，细节挺多，挺麻烦，但是很考验数学能力

不讲了，官方题解给的很清楚

Code:

%: pragma GCC optimize("O9")
#include<bits/stdc++.h>
#define ri register signed
#define p(i) ++i 
using namespace std;
namespace IO{
    char buf[1<<21],*p1=buf,*p2=buf,OPUT[100];
    #define gc() p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?(-1):*p1++
    template<typename T>inline void read(T &x) {
        ri f=1;x=0;register char ch=gc();
        while(!isdigit(ch)) {if (ch=='-') f=0;ch=gc();}
        while(isdigit(ch)) {x=(x<<1)+(x<<3)+(ch^48);ch=gc();}
        x=f?x:-x;
    } 
    template<typename T>inline void print(T x,char t) {
        if (x<0) putchar('-'),x=-x;
        if (!x) return putchar('0'),(void)putchar(t);     
        ri cnt(0);
        while(x) OPUT[p(cnt)]=x%10,x/=10;
        for (ri i(cnt);i;--i) putchar(OPUT[i]^48);
        return (void)putchar(t);   
    }
}
using IO::read;using IO::print;
namespace nanfeng{
    #define FI FILE *IN
    #define FO FILE *OUT
    #define int long long
    template<typename T>inline T cmax(T x,T y) {return x>y?x:y;}
    template<typename T>inline T cmin(T x,T y) {return x>y?y:x;}
    static const int MOD=3e5+7;
    int frac[MOD+7],inv[MOD+7],n,m,k,T,ans,bs1,bs2;
    inline int fpow(int x,int y) {
        int res(1);
        while(y) {
            if (y&1) res=res*x%MOD;
            x=x*x%MOD;
            y>>=1;
        }
        return res;
    }
    inline void init() {
        bs1=fpow(2,MOD-2),bs2=fpow(3,MOD-2);
        frac[1]=1;
        for (ri i(2);i<=MOD-1;p(i)) frac[i]=frac[i-1]*i%MOD;
        inv[MOD-1]=fpow(frac[MOD-1],MOD-2);
        for (ri i(MOD-2);i;--i) inv[i]=inv[i+1]*(i+1)%MOD;  
    } 
    inline int C(int n,int m) {
        if (n<m) return 0;
        if (n==m||!m) return 1;
        return frac[n]*inv[n-m]%MOD*inv[m]%MOD;
    }
    int lucas(int n,int m) {
        if (n<m) return 0;
        if (n==m||!m) return 1;
        return C(n%MOD,m%MOD)*lucas(n/MOD,m/MOD)%MOD;
    }
    inline int MD(int &x) {return x=x>=MOD?x-MOD:x;}
    inline int solve0() {
        int tmp1=(lucas(n,k)*(m%MOD)%MOD+lucas(m,k)*(n%MOD)%MOD)%MOD;
        int tmp2=(2*lucas(m+1,k+1)+2*lucas(m,k)*((n-m)%MOD)%MOD+2*lucas(m,k+1)%MOD)%MOD;
        return (tmp1+tmp2)>=MOD?(tmp1+tmp2)-MOD:tmp1+tmp2;
    }
    
    inline int calc1(int x) {return bs2*x*(x+1)%MOD*(x+bs1)%MOD;}
    inline int calc2(int x) {return (x*(x+1)>>1ll)%MOD;}
    inline int solve1() {return (n%MOD)*(m%MOD)%MOD;}
    inline int solve3() {
        int x=(cmin(n+1,(m+2)>>1)-1)%MOD;
        int y=(cmin(m+1,(n+2)>>1)-1)%MOD;
        n%=MOD,m%=MOD;
        int tmp1=2*(n*m*(x+y)%MOD+2*(calc1(x)+calc1(y))%MOD-(2*n+m)*calc2(x)%MOD-(2*m+n)*calc2(y)%MOD+MOD)%MOD;
        int tmp2=4*(n*m%MOD*(m-1)%MOD+calc1(m-1)-(n+m)*calc2(m-1)%MOD+MOD)%MOD;
        return (tmp1+tmp2+MOD)%MOD;
    }
    inline int solve4() {
        int x=(cmin(n+1,(m+2)>>1)-1)%MOD;
        int y=(cmin(m+1,(n+2)>>1)-1)%MOD;
        int z=(m>>1)%MOD;
        n%=MOD,m%=MOD;
        int tmp1=2*(n*m%MOD*(x+y)%MOD+2*(calc1(x)+calc1(y))%MOD-(2*n+m)*calc2(x)%MOD-(2*m+n)*calc2(y)%MOD+MOD)%MOD;
        int tmp2=(n*m%MOD*(m-1)%MOD+calc1(m-1)-(n+m)*calc2(m-1)%MOD+MOD)%MOD;
        int tmp3=5*(n*m*z%MOD-2*(n+m)*calc2(z)%MOD+4*calc1(z)+MOD)%MOD;
        return (tmp1+tmp2+tmp3+MOD)%MOD;
    }
    inline int solve5() {
        int x=(m-1>>1)%MOD;
        n%=MOD,m%=MOD;
        int tmp=(2*n*m*x%MOD+8*calc1(x)%MOD-4*(n+m)*calc2(x)%MOD+MOD)%MOD;
        return tmp;
    }
    inline int main() {
        //FI=freopen("nanfeng.in","r",stdin);
        //FO=freopen("nanfeng.out","w",stdout);
        init();
        read(T);
        for (ri z(1);z<=T;p(z)) {
            ans=0;
            read(n),read(m),read(k);
            if (n<m) swap(n,m);
            if (k>1) ans=solve0();
            switch(k) {
                case 1:ans=solve1(),MD(ans);break;
                case 3:ans+=solve3(),MD(ans);break;
                case 4:ans+=solve4(),MD(ans);break;
                case 5:ans+=solve5(),MD(ans);break;
                default:break;
            }
            print(ans,'\n');
        }
        return 0;
    }
    #undef int
}
int main() {return nanfeng::main();}