uva 10717 - Mint(欧几里得求最小公倍数）

来源：互联网发布：知乎海贼王路飞的回答编辑：程序博客网时间：2024/05/26 02:55

题目链接：uva 10717 - Mint

题目大意：给出n和m，然后给出n种硬币的的厚度，现在在给出m个桌子的高度，要求对应每个桌子high，输出两个值，从n种硬币中选取4种，每一种硬币用若干个叠在一起，使得四个柱一样吧高h,先在要找出一个h小于等于high，一个h大于等于high，两个值都要尽量接近high。

解题思路：枚举四种硬币，求出它们的最小公倍数。

#include <stdio.h>#include <string.h>#define min(a,b) (a)<(b)?(a):(b)#define max(a,b) (a)>(b)?(a):(b)const int N = 105;const int INF = 0x3f3f3f3f;int n, m;long long coin[N], high[N], l[N], r[N];void init() {for (int i = 0; i < n; i++)scanf("%lld", &coin[i]);for (int i = 0; i < m; i++)scanf("%lld", &high[i]);memset(l, 0, sizeof(l));memset(r, INF, sizeof(r));}long long gcd(long long a, long long b) {return b == 0 ? a : gcd(b, a % b);}void judge(long long c) {for (int i = 0; i < m; i++) {long long d = c;while (1) {if (d == high[i]) {l[i] = max(l[i], d);r[i] = min(r[i], d);break;}else if (d > high[i]) {l[i] = max(l[i], d - c);r[i] = min(r[i], d);break;}d += c;}}}void solve() {long long c, d;for (int i = 0; i < n; i++) {for (int j = i + 1; j < n; j++) {d = gcd(coin[i], coin[j]);d = coin[i] * coin[j] / d;for (int k = j + 1; k < n; k++) {for (int t = k + 1; t < n; t++) {c = gcd(coin[k], coin[t]);c = coin[k] * coin[t] / c;long long now = gcd(c, d);judge(c * d / now);}}}}}int main () {while (scanf("%d%d", &n, &m), n || m) {init();solve();for (int i = 0; i < m; i++)printf("%lld %lld\n", l[i], r[i]);}return 0;}