UVA 10717 Mint(多个数最小公倍数)

Problem E: Mint

The Royal Canadian Mint has commissioned a new series of designer coffee tables, with legs that are constructed from stacks of coins. Each table has four legs, each of which uses a different type of coin. For example, one leg might be a stack of quarters, another nickels, another loonies, and another twonies. Each leg must be exactly the same length.

Many coins are available for these tables, including foreign and special commemorative coins. Given an inventory of available coins and a desired table height, compute the lengths nearest to the desired height for which four legs of equal length may be constructed using a different coin for each leg.

Input consists of several test cases. Each case begins with an integers: 4 <= n<= 50 giving the number of types of coins available, and 1 <= t <= 10 giving the number of tables to be designed. n lines follow; each gives the thickness of a coin in hundredths of millimetres. t lines follow; each gives the height of a table to be designed (also in hundredths of millimetres). A line containing 0 0 follows the last test case.

For each table, output a line with two integers: the greatest leg length not exceeding the desired length, and the smallest leg length not less than the desired length.

Sample Input

4 250100200400100020000 0

Output for Sample Input

800 12002000 2000

题意：给定一些硬币的高度，要求用4种硬币组成桌子4条腿，要一样高，然后给定一个high，求出最接近high的两个数字（向下取和向上取)

思路：暴力枚举每4个数的最小公倍数。答案即为这些最小公倍数*k中。求法参考http://blog.163.com/ocean_123/blog/static/1908970672011621103949535/

代码：

#include <stdio.h>#include <string.h>const int MAXN = 55;int n, t, i, j, Min, Max;int h[MAXN], high, save[MAXN][MAXN][MAXN][MAXN];int max(int a, int b) {return a > b ? a : b;}int min(int a, int b) {return a < b ? a : b;}int gcd(int a, int b) {if (!b) return a;return gcd(b, a % b);}void sav() {for (int i = 0; i < n; i ++)for (int j = i + 1; j < n; j ++)for (int k = j + 1; k < n; k ++)for (int l = k + 1; l < n; l ++)if (!save[i][j][k][l]) {int mul = h[i] * h[j] * h[k] * h[l];save[i][j][k][l] = mul / gcd(mul / h[i], gcd(mul / h[j], gcd(mul / h[k], mul / h[l])));}}void solve() {for (int i = 0; i < n; i ++)for (int j = i + 1; j < n; j ++)for (int k = j + 1; k < n; k ++)for (int l = k + 1; l < n; l ++) {if (!(high % save[i][j][k][l])) {Max = Min = high;return;}else {Max = min(Max, (high / save[i][j][k][l] + 1) * save[i][j][k][l]);Min = max(Min, high / save[i][j][k][l] * save[i][j][k][l]);}}}int main() {while (~scanf("%d%d", &n, &t) && n || t) {memset(save, 0, sizeof(save));for (i = 0; i < n; i ++)scanf("%d", &h[i]);sav();while (t --) {scanf("%d", &high);Max = 999999999; Min = 0;solve();printf("%d %d\n", Min, Max);}}return 0;}