uva 10623 - Thinking Backward(数学)

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
;

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;
}