题目链接：uva 10483 - The Sum Equals the Product
题目大意：给出n和m，然后找出a，b，c，sum = a + b + c = a * b * c; n <= sum <= m。按照字典序输出。
解题思路：暴力枚举a和b，a <= pow(m, 1/3), a <= b <= sqrt(m/a).
#include#include #include #include using namespace std; const double tmp = 100.0; const double eps = 1e-9; const int N = 10005; struct state { double s, a, b, c; }ans[N]; int cnt; double n, m; bool cmp(const state& p, const state& q) { if (fabs(p.s - q.s) > eps) return p.s < q.s; if (fabs(p.a - q.a) > eps) return p.a < q.a; if (fabs(p.b - q.b) > eps) return p.b < q.b; return p.c < q.c; } void solve() { int N = n * 100, M = m * 100; for (int i = 1; i * i * i <= M * 10000; i++) { for (int j = i; j * j * i<= M * 10000; j++) { if (i * j <= 10000 || (i + j) * 10000 % (i * j - 10000)) continue; int x = (i + j) * 10000 / (i * j - 10000); if (x < j || x < i) continue; int sum = i + j + x; int f = i * j * x; if (f % 10000) continue; if (sum >= N && M >= sum && sum == f / 10000) { ans[cnt].s = sum/tmp; ans[cnt].a = i/tmp; ans[cnt].b = j/tmp; ans[cnt].c = x/tmp; cnt++; } } } sort(ans, ans + cnt, cmp); for (int i = 0; i < cnt; i++) { printf("%.2lf = %.2lf + %.2lf + %.2lf = %.2lf * %.2lf * %.2lf\n", ans[i].s, ans[i].a, ans[i].b, ans[i].c, ans[i].a, ans[i].b, ans[i].c); } } int main() { while (scanf("%lf%lf", &n, &m) == 2) { cnt = 0; solve(); } return 0; }