UVa 11401 Triangle Counting (计数DP)

题意：给定一个数 n，从1-n这些数中任意挑出3个数，能组成三角形的数目。

析：dp[i] 表示从1-i 个中任意挑出3个数，能组成三角形的数目。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e6 + 5;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline int gcd(int a, int b){ return b ? gcd(b, a%b) : a; }
inline int lcm(int a, int b){ return a * b / gcd(a, b); }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
LL dp[maxn];

int main(){
    dp[3] = 0;
    for(int i = 4; i < maxn; ++i){
        if(i & 1)  dp[i] = dp[i-1] + (LL)(i-1)*((i-4)/2+1)/2;
        else dp[i] = dp[i-1] + (LL)((i-3)/2+1)*(i-2)/2;
    }
    while(cin >> n && n > 2){
        cout << dp[n] << endl;
    }
    return 0;
}