UVA 11401 Triangle Counting [递推]-优快云博客

UVA 11401 Triangle Counting [递推]

题目链接：VJudge
题意：给定N条边，边长分别为1~N，从这N条边中，选出三条边，问能构成三角形的情况有多少种。
思路：dp[i]表示i条边的情况，dp[i]包含了dp[i-1]与最长边为i这两类情况构成。当最长边为i时，另外两条边范围是[2, i-1],我们假定这两条边为a,b。满足条件1

#include <cmath>
#include <queue>
#include <vector>
#include <cstdio>
#include <string>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
using namespace std;

//#pragma comment(linker, "/STACK:1024000000,1024000000")

#define FIN             freopen("input.txt","r",stdin)
#define FOUT            freopen("output.txt","w",stdout)
#define fst             first
#define snd             second
#define lson            l, mid, rt << 1
#define rson            mid + 1, r, rt << 1 | 1

// typedef __int64  LL;
typedef long long LL;
typedef unsigned int uint;

const int INF = 0x3f3f3f3f;
const double eps = 1e-6;
const int MAXN = 1000000 + 5;
const int MAXM = 10000 + 5;

LL dp[MAXN];
int N;

void init() {
    dp[3] = 0;
    for (uLL i = 4; i < MAXN; i++) {
        uLL t = (i - 2) >> 1, x;
        if (i & 1) {
            x = (1 + t) * t;
        } else {
            x = t * t;
        }
        dp[i] = dp[i - 1] + x;
    }
}

int main() {
#ifndef ONLINE_JUDGE
    FIN;
#endif // ONLINE_JUDGE
    init();
    while (~scanf("%d", &N) && N >= 3) {
        printf("%lld\n", dp[N]);
    }
    return 0;
}