题目传送：UVA – 10717
思路：思路很明确，就是找出所有的四种硬币的组合，然后求出他们的最小公倍数，再去找出该最小公倍数的倍数中与desired length最接近的，昨晚不知道啥错误，一直WA，今天重写一下就过了
AC代码：
#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <cmath>#include <queue>#include <stack>#include <vector>#include <map>#include <set>#include <deque>#include <cctype>#define LL long long#define INF 0x7fffffffusing namespace std;int n, t; int a[105];int L, R;int gcd(int a, int b) {return b == 0 ? a : gcd(b, a % b);}int lcm(int a, int b) {return a / gcd(a, b) * b;}void fun(int des) {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 ++) {int Lcm = lcm(a[i], lcm(a[j], lcm(a[k], a[l])));int num = des / Lcm;if(Lcm * num == des) {L = des; R = des; return;}else {L = max(L, num * Lcm);R = min(R, (num + 1) * Lcm);}}}}}}int main() {while(scanf("%d %d", &n, &t), n || t) {for(int i = 0; i < n; i ++) {scanf("%d", &a[i]);}for(int i = 0; i < t; i ++) {int des;scanf("%d", &des);L = – INF;R = INF;fun(des);printf("%d %d\n", L, R);}}return 0;}
，没有创造的生活不能算生活，只能算活着。
