SRM 552 div2

250pt water question

 

500pt

In two cases, 1. The minimum value in RGB is large enough. The constraint is the value of R + G + B. In this case, the result is (R + G + B) /tot 2, R + G + B is large enough, and the constraint is that the minimum value in RGB is not large enough. In this case, the result is that M/C takes the minimum value.

In the first case, r = 7, G = 6, B = 6. The minimum value of 6 is enough to draw two triangles, but R + G + B = 19 is not enough to draw two triangles.

In the second case, r = 1000000000, G = 6, B = 6, r + G + B is enough to draw n multiple triangles, but G and B have only 6 triangles. A maximum of two triangles can be drawn.

/* Six lines of code from his sister !!! */Class foxpaintingbils {public: long themax (long R, long g, long B, int N) {ll T = LL (N) * LL (n + 1)/2; If (n = 1) return r + G + B; ll c = T/3; ll x = min (r, min (G, B); LL ans = min (R + G + B)/t, x/C); Return ans ;}};

 

 

1000 PT

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