UVa 10717 Mint（LCM）

题意：

给定不同厚度的硬币n 选出其中的四个，然后组成四个桌腿并且要使每个桌腿高度一样

给定的高度h, 分别求不大于h的最大的高度和不小于h的最小高度

思路：

枚举硬币，然后更新状态即可。

#include <cstdio> #include <cstring> #include <cstring>



#define LL long long int 



int coin[60], tab[60]; LL low[20], high[20]; int n, t; LL gcd(LL x, LL y) { LL t; while (y) { t = x % y; x = y; y = t; } return x; } LL calclcm(LL a, LL b, LL c, LL d) { LL temp, ans; temp = gcd(a, b); ans = a * b / temp; temp = gcd(ans, c); ans = ans * c / temp; temp = gcd(ans, d); ans = ans * d / temp; return ans; } void solve() { memset(low, 0, sizeof(low)); memset(high, 0x3f, sizeof(high)); for (int i = 0; i < n; ++i) for (int j = i + 1; j < n; ++j) for (int p = j + 1; p < n; ++p) for (int q = p + 1; q < n; ++q) { LL ans = calclcm(coin[i], coin[j], coin[p], coin[q]); for (int k = 0; k < t; ++k) { if (ans < tab[k] && tab[k] - tab[k]%ans > low[k]) low[k] = tab[k] - tab[k]%ans; if ((ans - tab[k]%ans) % ans + tab[k] < high[k]) high[k] = (ans - tab[k]%ans) % ans + tab[k]; } } } int main() { while (scanf("%d %d", &n, &t) && n && t) { for (int i = 0; i < n; ++i) scanf("%d", &coin[i]); for (int i = 0; i < t; ++i) scanf("%d", &tab[i]); solve(); for (int i = 0; i < t; ++i) printf("%lld %lld\n", low[i], high[i]); } return 0; }