题目链接：uva 10623 - Thinking Backward

题目大意：就是给出N，表示要将平面分解成N份，问有哪些可选则的方案，m表示椭圆、n表示圆形、p表示三角形的个数，m、n、p分别给定范围。

解题思路：本来这题一点思路都没有，但是在论坛上看到一个公式N=2+2m(m?1)+n(n?1)+4mn+3p(p?1)+6mp+6np

这样只要枚举m和p，求解n，判断n是否满足即可，注意n一定是整数。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>

using namespace std;
typedef  long long ll;
const int N = 100005;

struct state {
    ll n, m, p;
    void set (ll m, ll n, ll p) {
        this->m = m;
        this->n = n;
        this->p = p;
    }
}s[N];

bool cmp (const state& a,const state& b) {
    if (a.m != b.m)
        return a.m < b.m;
    if (a.n != b.n)
        return a.n < b.n;
    return a.p < b.p;
}

int main () {
    int cas = 1;
    ll n;
    while (scanf("%lld", &n) == 1 && n != -1) {
        printf("Case %d:\n", cas++);
        if (n == 1) {
            printf("0 0 0\n");
            continue;
        }
        int c = 0;

        for (ll m = 0; m < 100; m++) {
            for (ll p = 0; p < 100; p++) {
                ll sum = 2 + 2 * m * (m-1) + 3 * p * (p-1) + 6 * m * p;
                sum = n - sum;

                ll a = 4 * m + 6 * p - 1;

                /*
                if (m == 0 && p == 0)
                    printf("%lld %lld! \n", sum, a);
                    */

                double tmp = sum + a * a / 4.0;



                if (tmp < 0)
                    continue;

                tmp = sqrt(tmp);
                double x = (tmp - ((double)a / 2.0));

                if (x < 0 || x >= 20000)
                    continue;

                ll n = x;
                /*
                    */
                if (n * n + a * n == sum)
                    s[c++].set(m, n, p);

            }
        }

        sort (s, s + c, cmp);
        if (c) {
            for (int i = 0; i < c; i++)
                printf("%lld %lld %lld\n", s[i].m, s[i].n, s[i].p);
        } else
            printf("Impossible.\n");
    }
    return 0;
}