HDU 12,493 Corner-shaped Division

The three edges of a triangle can be added to a graph to observe

Suppose you add the nth triangle

The first n-1 triangle divides the area into sum[n-1]

The nth triangle can pass up to two sides of each triangle on each edge of the first n-1 triangle, and one edge is cut to increase the n-1-1 area.

Then the three edges of the internal graph are cut to increase the 6* (n-1)-3 regions, and the new triangle itself in three corners of the formation of three new areas

The 6* (n-1) area has been added to a total of

Then the recursive function is

Sum[i] = sum[i-1] + 6* (i-1)

In fact, the direct point is to use the Euler formula to solve the problem

V (Point)-E (Edge) + F (face) = 2

Add nth triangles at a time

The added point is 2 * (n-1) * 3 + 3 = 6*n-3

Added side for (2*n-1) * * (new triangle added edge) + n-1 (original n-1 triangle each three edges are divided by the new triangle two times) = 6*n-9

So the last increase in the number of polygons is 6* (n-1)

1#include <cstdio>2 3 intsum[10005];4 5 intMain ()6 {7sum[1] =2;8      for(inti =2; i<=10000; i++)9Sum[i] = sum[i-1] +3*2* (I-1);Ten     intT, N; Onescanf"%d", &T); A      while(t--){ -scanf"%d", &n); -printf"%d\n", Sum[n]); the     } -     return 0; -}

HDU 12,493 Corner-shaped Division