Submission #19261

#	Time	Username	Problem	Language	Result	Execution time	Memory
19261		kaTkaHr	흑백 (kriii4_G)	C++14	6 / 100	2000 ms	1136 KiB

G

#include <stdio.h>
#include<vector>
#include <algorithm>
#include <map>

using namespace std;

typedef long long ll;

const int MX = 3005, MM = 1000000007;

ll pw(ll A, ll B){
    ll R = 1;
    while(B){
        if( B&1 ) R = R * A % MM;
        A = A * A % MM; B /= 2;
    }
    return R;
}


ll rv(ll A){ return pw(A, MM-2); }

struct frac{
    ll A, B;
		frac(ll A):A(A), B(1){}
    frac(ll a, ll b){
        A = (a%MM+MM) % MM;
        B = (b%MM+MM) % MM;
    }
    frac(){A = 0, B = 1;}
    frac operator+ (const frac &l)const{
        return frac((A * l.B + B * l.A) % MM, B * l.B % MM);
    }
    frac operator*(const frac &l)const{
        return frac(A*l.A % MM, B*l.B % MM);
    }
    frac operator/(const frac &l)const{
        return frac(A*l.B % MM, B*l.A % MM);
    }
    frac operator- (const frac &l)const{
        return frac((A*l.B - B*l.A%MM + MM) % MM, B*l.B % MM);
    }
    ll v(){ return A * rv(B) % MM; }
};

frac pw(frac A, ll B){
    frac R = 1;
    while(B){
        if( B&1 ) R = R * A;
        A = A * A; B /= 2;
    }
    return R;
}

frac F[MX];

int main()
{
	F[0] = 1;
	for(int i = 1; i <= 1000; i++) F[i] = F[i-1] * i;
	int H, W;
	frac _5 = frac(1, 2);
	scanf("%d%d", &H, &W);
	frac A = 0, B = 0, C = 0;
	for(int i = 1; i <= W; i++){
		for(int j = 1; j <= W-i; j++){
			int k = W-i-j;
			A = A + F[W] / F[i] / F[j] / F[k] * (
				pw(pw(_5, i) + pw(_5, j) + 1, H) 
				- pw(pw(_5, i) + 1, H) 
				- pw(pw(_5, j) + 1, H) 
				+ 1
			);
		}
	}
	
	for(int i = 1; i <= W; i++){
		for(int j = 1; j <= W-i; j++){
			int k = W-i-j;
			B = B + F[W] / F[i] / F[j] / F[k] * (
				pw(pw(_5, i) + pw(_5, j) + pw(_5,i+j) + 1, H) 
				- pw(pw(_5, i) + pw(_5, j) + 1, H) 
			);
		}
	}
	for(int i = 1; i <= H; i++){
		for(int j = 1; j <= H-i; j++){
			int k = H-i-j;
			C = C + F[H] / F[i] / F[j] / F[k] * (
				pw(pw(_5, i) + pw(_5, j) + pw(_5,i+j) + 1, W) 
				- pw(pw(_5, i) + pw(_5, j) + 1, W) 
			);
		}
	}
	frac ans = B + C + A;
	printf("%lld\n", ans.v());
}

• Subtask 1: 6 / 6
• Subtask 2: 0 / 94