Codechef Not a Triangle

Find the three numbers in an array. The three numbers cannot form triangles.

The condition that the three numbers cannot form a triangle is: a + B <c

And smaller than the third side.

This problem is a combination of three numbers. The violence law can be solved, but the time efficiency is O (n * n), which is very slow.

However, since it is a combination problem, the post-sorting method can be used to reduce the time efficiency.

The method to reduce the time efficiency is as follows:

Select a maximum number c, and then select two decimals a and B, where a <B. if the conditions are met, then, the numbers in the two numbers must be used as B to meet the condition a + B <c

In this way, the time efficiency can be reduced to O (n * n ).

This rule is hard to find. Take a very careful look and summarize it.

To process an array, We need to observe the rule from small to large, from large to small.

Original question: http://www.codechef.com/problems/NOTATRI/

# Include
 
  
# Include
  
   
# Include using std: sort; class NotATriangle {// carefully find out the rule to optimize int getNotTris_3 (int * A, const int n) {sort (A, A + n ); int I = 0, j = 1, k = n-1, ans = 0; for (; k> = 2; k --) {I = 0, j = K-1; while (I <j) {if (A [I] + A [j] <A [k]) {ans + = j-I; I ++ ;} else j -- ;}} return ans;} public: NotATriangle () {int N = 0; while (scanf ("% d", & N) & 0! = N) {int * A = (int *) malloc (sizeof (int) * N); for (int I = 0; I <N; I ++) {scanf ("% d", & A [I]);} printf ("% d \ n", getNotTris_3 (A, N )); free (A) ;}}; int notATriangle () {NotATriangle ntio; return 0 ;}