1 条题解

  • 0
    @ 2023-6-21 20:03:30

    C++ :

    #define _GLIBCXX_IOSTREAM
    #include<bits/stdc++.h>
    
    typedef unsigned int uint;
    typedef long long ll;
    typedef unsigned long long ull;
    typedef double lf;
    typedef long double llf;
    typedef std::pair<int,int> pii;
    
    #define xx first
    #define yy second
    
    template<typename T> inline T max(T a,T b){return a>b?a:b;}
    template<typename T> inline T min(T a,T b){return a<b?a:b;}
    template<typename T> inline T abs(T a){return a>0?a:-a;}
    template<typename T> inline bool repr(T &a,T b){return a<b?a=b,1:0;}
    template<typename T> inline bool repl(T &a,T b){return a>b?a=b,1:0;}
    template<typename T> inline T gcd(T a,T b){T t;if(a<b){while(a){t=a;a=b%a;b=t;}return b;}else{while(b){t=b;b=a%b;a=t;}return a;}}
    template<typename T> inline T sqr(T x){return x*x;}
    #define mp(a,b) std::make_pair(a,b)
    #define pb push_back
    #define I inline
    #define mset(a,b) memset(a,b,sizeof(a))
    #define mcpy(a,b) memcpy(a,b,sizeof(a))
    
    #define fo0(i,n) for(int i=0,i##end=n;i<i##end;i++)
    #define fo1(i,n) for(int i=1,i##end=n;i<=i##end;i++)
    #define fo(i,a,b) for(int i=a,i##end=b;i<=i##end;i++)
    #define fd0(i,n) for(int i=(n)-1;~i;i--)
    #define fd1(i,n) for(int i=n;i;i--)
    #define fd(i,a,b) for(int i=a,i##end=b;i>=i##end;i--)
    #define foe(i,x)for(__typeof(x.end())i=x.begin();i!=x.end();++i)
    
    struct Cg{I char operator()(){return getchar();}};
    struct Cp{I void operator()(char x){putchar(x);}};
    #define OP operator
    #define RT return *this;
    #define RX x=0;char t=P();while((t<'0'||t>'9')&&t!='-')t=P();bool f=0;\
    if(t=='-')t=P(),f=1;x=t-'0';for(t=P();t>='0'&&t<='9';t=P())x=x*10+t-'0'
    #define RL if(t=='.'){lf u=0.1;for(t=P();t>='0'&&t<='9';t=P(),u*=0.1)x+=u*(t-'0');}if(f)x=-x
    #define RU x=0;char t=P();while(t<'0'||t>'9')t=P();x=t-'0';for(t=P();t>='0'&&t<='9';t=P())x=x*10+t-'0'
    #define TR *this,x;return x;
    I bool IS(char x){return x==10||x==13||x==' ';}template<typename T>struct Fr{T P;I Fr&OP,(int &x)
    {RX;if(f)x=-x;RT}I OP int(){int x;TR}I Fr&OP,(ll &x){RX;if(f)x=-x;RT}I OP ll(){ll x;TR}I Fr&OP,(char &x)
    {for(x=P();IS(x);x=P());RT}I OP char(){char x;TR}I Fr&OP,(char *x){char t=P();for(;IS(t);t=P());if(~t){for(;!IS
    (t)&&~t;t=P())*x++=t;}*x++=0;RT}I Fr&OP,(lf &x){RX;RL;RT}I OP lf(){lf x;TR}I Fr&OP,(llf &x){RX;RL;RT}I OP llf()
    {llf x;TR}I Fr&OP,(uint &x){RU;RT}I OP uint(){uint x;TR}I Fr&OP,(ull &x){RU;RT}I OP ull(){ull x;TR}};Fr<Cg>in;
    #define WI(S) if(x){if(x<0)P('-'),x=-x;char s[S],c=0;while(x)s[c++]=x%10+'0',x/=10;while(c--)P(s[c]);}else P('0')
    #define WL if(y){lf t=0.5;for(int i=y;i--;)t*=0.1;if(x>=0)x+=t;else x-=t,P('-');*this,(ll)(abs(x));P('.');if(x<0)\
    x=-x;while(y--){x*=10;x-=floor(x*0.1)*10;P(((int)x)%10+'0');}}else if(x>=0)*this,(ll)(x+0.5);else *this,(ll)(x-0.5);
    #define WU(S) if(x){char s[S],c=0;while(x)s[c++]=x%10+'0',x/=10;while(c--)P(s[c]);}else P('0')
    template<typename T>struct Fw{T P;I Fw&OP,(int x){WI(10);RT}I Fw&OP()(int x){WI(10);RT}I Fw&OP,(uint x){WU(10);RT}
    I Fw&OP()(uint x){WU(10);RT}I Fw&OP,(ll x){WI(19);RT}I Fw&OP()(ll x){WI(19);RT}I Fw&OP,(ull x){WU(20);RT}I Fw&OP()
    (ull x){WU(20);RT}I Fw&OP,(char x){P(x);RT}I Fw&OP()(char x){P(x);RT}I Fw&OP,(const char *x){while(*x)P(*x++);RT}
    I Fw&OP()(const char *x){while(*x)P(*x++);RT}I Fw&OP()(lf x,int y){WL;RT}I Fw&OP()(llf x,int y){WL;RT}};Fw<Cp>out;
    
    const int n2=512,n5=1953125;
    
    int m2[n2],m5[n5];
    
    void exgcd(int a,int b,int&x,int&y)
    {
    	if(b)exgcd(b,a%b,y,x),y-=(a/b)*x;
    	else x=1,y=0;
    }
    
    inline int inv(int a,int b)
    {
    	int x,y;
    	exgcd(b,a,x,y);
    	return y<0?y+b:y;
    }
    
    template<typename T>inline int pow(int a,T b,int c)
    {
    	int r=1;
    	for(;b;(b>>=1)?a=(ll)a*a%c:0)
    		if(b&1)r=(ll)r*a%c;
    	return r;
    }
    
    struct data
    {
    	int x;ll y;
    	inline void mul(const data&v,int m)
    	{
    		x=(ll)x*v.x%m,y+=v.y;
    	}
    	inline void div(const data&v,int m)
    	{
    		x=(ll)x*inv(v.x,m)%m,y-=v.y;
    	}
    };
    
    data g2(ll x)
    {
    	data ret=(data){pow(m2[n2-1],x/n2,n2)*m2[x%n2]%n2,x/2};
    	if(x<2)return ret;
    	ret.mul(g2(x/2),n2);
    	return ret;
    }
    
    data g5(ll x)
    {
    	data ret=(data){(int)((ll)pow(m5[n5-1],x/n5,n5)*m5[x%n5]%n5),x/5};
    	if(x<5)return ret;
    	ret.mul(g5(x/5),n5);
    	return ret;
    }
    
    inline int c2(ll n,ll m)
    {
    	data a=g2(n);
    	a.div(g2(m),n2);
    	a.div(g2(n-m),n2);
    	a.x=a.x*pow(2,a.y,n2)%n2;
    	return a.x;
    }
    
    inline int c5(ll n,ll m)
    {
    	data a=g5(n);
    	a.div(g5(m),n5);
    	a.div(g5(n-m),n5);
    	a.x=(ll)a.x*pow(5,a.y,n5)%n5;
    	return a.x;
    }
    
    inline int C(ll n,ll m)
    {
    	int a=c2(n,m),b=c5(n,m);
    	//out,n,' ',m,' ',a,' ',b,' ',(787109376ll*b+212890625ll*a)%1000000000,'\n';
    	return(787109376ll*b+212890625ll*a)%1000000000;
    }
    
    inline int d2(ll n,ll m)
    {
    	data a=g2(n);
    	a.div(g2(m),n2);
    	a.div(g2(n-m),n2);
    	a.x=a.x*pow(2,a.y-1,n2)%n2;
    	return a.x;
    }
    
    inline int d5(ll n,ll m)
    {
    	data a=g5(n);
    	a.div(g5(m),n5);
    	a.div(g5(n-m),n5);
    	a.x=(ll)a.x*pow(5,a.y,n5)%n5;
    	return (ll)a.x*(n5+1)/2%n5;
    }
    
    inline int D(ll n,ll m)
    {
    	int a=d2(n,m),b=d5(n,m);
    	//out,n,' ',m,' ',a,' ',b,' ',(787109376ll*b+212890625ll*a)%1000000000,'\n';
    	return(787109376ll*b+212890625ll*a)%1000000000;
    }
    
    int main()
    {
    	m2[0]=1;
    	for(int i=1;i<n2;i++)
    	{
    		if(i%2==0)m2[i]=m2[i-1];
    		else m2[i]=m2[i-1]*i%n2;
    	}
    	m5[0]=1;
    	for(int i=1;i<n5;i++)
    	{
    		if(i%5==0)m5[i]=m5[i-1];
    		else m5[i]=(ll)m5[i-1]*i%n5;
    	}
    	int k;ll n,m;
    	while(~scanf("%lld%lld%d",&n,&m,&k))
    	{
    		int c=pow(10,k,1000000007);
    		char s[6];sprintf(s,"%%0%dd\n",k);
    		int ans=pow(2,n+m-1,c);
    		//out,n,' ',m,' ',c,' ',ans,'\n';
    		if((n+m)%2==0)
    		{
    			if(m<n)(ans+=D(n+m,(n+m)/2))%=c;
    			else(ans-=D(n+m,(n+m)/2))%=c;
    			//out,"naive\n";
    		}
    		for(ll i=(n+m-1)/2;i>m;i--)
    			(ans+=C(n+m,i))%=c;
    		(ans+=c)%=c;
    		printf(s,ans);
    	}
    }
    
    • 1

    信息

    ID
    1022
    时间
    10000ms
    内存
    128MiB
    难度
    (无)
    标签
    (无)
    递交数
    0
    已通过
    0
    上传者