UVa 10483 - The Sum Equals the Product (枚举)
题意
找出一个数,使得$sum = a + b + c = a * b * c$
思路
看的帆神。
枚举a和b,求出c,a和b在算的过程中可以剪枝一下。
代码
#include <cstdio>
#include <stack>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <iomanip>
#include <algorithm>
#include <cctype>
#include <string>
#include <map>
#include <cmath>
using namespace std;
#define LL long long
#define ULL unsigned long long
#define SZ(x) (int)x.size()
#define Lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define F first
#define S second
#define ROP freopen("input.txt", "r", stdin);
#define MID(a, b) (a + ((b - a) >> 1))
#define LC rt << 1, l, mid
#define RC rt << 1|1, mid + 1, r
#define LRT rt << 1
#define RRT rt << 1|1
#define BitCount(x) __builtin_popcount(x)
#define BitCountll(x) __builtin_popcountll(x)
#define LeftPos(x) 32 - __builtin_clz(x) - 1
#define LeftPosll(x) 64 - __builtin_clzll(x) - 1
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const double eps = 1e-8;
const int MAXN = 100 + 10;
const int MOD = 1000007;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
typedef pair<int, int> pii;
typedef vector<int>::iterator viti;
typedef vector<pii>::iterator vitii;
struct EQUATION
{
double sum, a, b, c;
};
vector<EQUATION> ans;
int nn, mm;
bool cmp(const EQUATION &a, const EQUATION &b)
{
if (fabs(a.sum - b.sum) >= eps) return a.sum < b.sum;
if (fabs(a.a - b.a) >= eps) return a.a < b.a;
if (fabs(a.b - b.b) >= eps) return a.b < b.b;
if (fabs(a.c - b.c) >= eps) return a.c < b.c;
}
bool Judge(int a, int b, int c)
{
if (c < a || c < b) return false;
int x = a + b + c, y = a * b * c;
if (x < nn || x > mm) return false;
if (y % 10000) return false;
if (x == y / 10000) return true;
else return false;
}
int main()
{
//ROP;
int i, j;
double n, m;
while (cin >> n >> m)
{
ans.clear();
nn = (int)(n * 100 + 0.5), mm = (int)(m * 100 + 0.5);
for (i = 1; i * i * i <= mm * 10000; i++)
{
for (j = i; i * j * j <= mm * 10000; j++)
{
int a = i + j, b = i * j;
if (b <= 10000 || 10000 * (i + j) % (i * j - 10000)) continue;
int k = 10000 * (i + j) / (i * j - 10000);
if (!Judge(i, j, k)) continue;
else
{
EQUATION tmp;
tmp.sum = (i + j + k) * 1.0 / 100;
tmp.a = i * 1.0 / 100; tmp.b = j * 1.0 / 100; tmp.c = k * 1.0 / 100;
ans.PB(tmp);
}
}
}
sort(ans.begin(), ans.end(), cmp);
for (i = 0; i < SZ(ans); i++) printf("%.2f = %.2f + %.2f + %.2f = %.2f * %.2f * %.2f\n",
ans[i].sum, ans[i].a, ans[i].b, ans[i].c, ans[i].a,
ans[i].b, ans[i].c);
}
return 0;
}