UVA11401 Triangle Counting

Triangle Counting

 

题目大意：给定n,n >= 3,求用1,2,3,...,n-1,n这些木棒能拼出的三角形个数

 

 

尝试设三条边为x,y,z，发现仅通过x + y < z, x - y < z无法找到解题方向

不妨再加一些附加条件，设x为最大边长度，就有x - y < z < x,只需计算此时x,y,z取值个数，加和即可

y = 1时无解；y = 2时z = x - 1;y = 3时z = x - 1, x - 2;....y = i时，z有i - 1种取值

共有Σ(i = 0 to x - 2)i种取值，即(n - 2) * (n - 1) / 2

不难发现存在z = y的情况，去掉即可，即y > x/2时，z = x，于是有x - 1 - ([x / 2] + 1) + 1 =  x - [x / 2] -  1 = [(x + 1) / 2] - 1 = [(x - 1) / 2]

每个三角形考虑了两边，最后答案/2

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <vector>
#define min(a, b) ((a) < (b) ? (a) : (b))
#define max(a, b) ((a) > (b) ? (a) : (b))
#define abs(a) ((a) < 0 ? (-1 * (a)) : (a))
inline void swap(long long &a, long long &b)
{
    long long tmp = a;a = b;b = tmp;
}
inline void read(long long &x)
{
    x = 0;char ch = getchar(), c = ch;
    while(ch < '0' || ch > '9') c = ch, ch = getchar();
    while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
    if(c == '-') x = -x;
}

const long long INF = 0x3f3f3f3f;
const long long MAXN = 1000000 + 10;

long long f[MAXN], n;

int main()
{
    for(register long long i = 1;i <= 1000000;++ i)
        f[i] = f[i - 1] + ((((i - 2) * (i - 1) >> 1) - ((i - 1) >> 1)) >> 1);
    while(scanf("%lld", &n) != EOF && n >= 3)
    {
        printf("%lld\n", f[n]);
    }
    return 0;
}