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Create Knight Explanation.txt

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	`1`	`+Firstly, the constraints are too big to do a DP.`
	`2`	`+`
	`3`	`+1. Let us notice the sum (i, j) (mod 3) is an invariant at each step.`
	`4`	`+`
	`5`	`+Initially it is 0. So, if (X + Y) = 0 (mod 3), we can reach (X, Y). Otherwise, no.`
	`6`	`+`
	`7`	`+2. Now, let us suppose that we make R moves of the kind (i + 1, j + 2) and D moves of the kind (i + 2, j + 1).`
	`8`	`+`
	`9`	`+Then,`
	`10`	`+`
	`11`	`+R + 2D = X`
	`12`	`+2R + D = Y`
	`13`	`+`
	`14`	`+Since we know the values of X and Y, let us try to write (D, R) in terms of (X, Y)`
	`15`	`+`
	`16`	`+2X - Y = 3D`
	`17`	`+2Y - X = 3R`
	`18`	`+`
	`19`	`+3. Now that we know the values of (D, R), we need to count the number of strings of length (D + R) with D d's and R r's. This string is uniquely determined by the placement of the d's or the r's.`
	`20`	`+`
	`21`	`+So, the answer = C(D + R, D) = C(D + R, R)`
	`22`	`+`
	`23`	`+Or course, both D and R have to be > 0`
	`24`	`+`
	`25`	`+-----`
	`26`	`+`
	`27`	`+int main()`
	`28`	`+{`
	`29`	`+ int X, Y;`
	`30`	`+ cin >> X >> Y;`
	`31`	`+`
	`32`	`+ if( (X + Y)%3 != 0)`
	`33`	`+ {`
	`34`	`+ cout << "0\n";`
	`35`	`+`
	`36`	`+ return 0;`
	`37`	`+ }`
	`38`	`+`
	`39`	`+ long long D = (2*Y - X)/3;`
	`40`	`+ long long L = (2*X - Y)/3;`
	`41`	`+`
	`42`	`+ const int MOD = 1e9 + 7;`
	`43`	`+ cout << ( (D < 0 \|\| L < 0) ? 0 : combinations(D + L, D, MOD) );`
	`44`	`+`
	`45`	`+ return 0;`
	`46`	`+}`
	`47`	`+`

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