uva 10623 - Thinking Backward(數學)

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題目連結：uva 10623 - Thinking Backward

題目大意：就是給出N，表示要將平面分解成N份，問有哪些可選則的方案，m表示橢圓、n表示圓形、p表示三角形的個數，m、n、p分別給定範圍。

解題思路：本來這題一點思路都沒有，但是在論壇上看到一個公式N=2+2m(m?1)+n(n?1)+4mn+3p(p?1)+6mp+6np

這樣只要枚舉m和p，求解n，判斷n是否滿足即可，注意n一定是整數。

#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;}