UVA Triangle Counting 11401 "geometry + math"

11401-triangle countingtime limit:1.000 seconds

Test instructions: give you n a line, length 1-n. Ask how many different triangles can be formed.

Problem Solving Ideas:

The triangles with the largest edge length x have C (x), the other two sides are Y and Z respectively, and have y+z>x according to the triangular inequalities. So Z's range is x-y < Z < x.

① According to this inequality, when Y=1 x-1 < Z < x, there is no solution, y=2 there is a solution (z=x-1), y=3, there are two solutions (z=x-1 or z=x-2) ... Until Y=x-1, there is a x-2 solution. According to the arithmetic progression summation formula, there are altogether 0+1+2+......+ (x-3) + (x-2) = (x-1) (x-2)/2 solutions.

The ② above contains the Y=z case, and each triangle is two times. So we should count the y=z of the situation. When y = Z is x-z < Z < x, then there is x < 2z, X/2 < Z < x, so the number of Z (x-1) is met (x/2+1)

Then subtract the solutions and divide by 2.

Why does ③ calculate both sides? You can push the x=5 y case starting from 1. You will see that the triangles are repeated two times.

C (x) = ((i-2) * (i-1)/2-((i-1)-(i/2+1))/2;; The original question required "the number of triangles with a maximum edge length of not more than n" F (n) ", then f (n) =c (1) +c (2) +...+c (n). Written recursively is f (n) = f (n-1) + C (n).

AC Code:

#include <stdio.h> #include <math.h> #include <vector> #include <queue> #include <string> #include <string.h> #include <stdlib.h> #include <iostream> #include <algorithm>using Namespace Std;typedef long Long ll;const int maxn = 1000002; LL F[maxn];int Main () {    f[3]=0;    for (LL i=4;i<maxn;i++)        f[i]=f[i-1]+ ((i-2) * (i-1)/2-((i-1)-(i/2+1))/2;    int n;    while (scanf ("%d", &n)} {        if (n<3) break;        printf ("%lld\n", F[n]);    }    return 0;}

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UVA Triangle Counting 11401 "geometry + math"