coins.zip分治算法，求取最轻且价值最大的硬币集合

动态规划，要求：现有硬币n枚。其价值为v(1,q,q^2……q^n),且每枚硬币重量为一，求价值为Y且重量最小的硬币集合 时间复杂度为O(n*v)

#include
using namespace std;
#define P 3
#define N 15
#define D 1800
int main（） {

	int distance[N];
	distance[0] = 1;
	for （int i = 1; i < N; i++）
		distance[i] = distance[i - 1] * P;

	int barrel = 0; int number = 0; int flag[N];
	int j = 0;
	for （int i = 0; i < N; i++） {
		if （barrel + distance[i] <= D） {
			barrel += distance[i];
		}
		else {
			flag[j++] = i;
			barrel = 0;
		}
	}

		
	
	system（“pause“）;
}

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     目录           0  2019-05-09 17:52  Algorithim_lab03_02\.vs\
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     目录           0  2019-05-09 17:52  Algorithim_lab03_02\Algorithim_lab03_02\
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     目录           0  2019-05-09 17:52  Algorithim_lab03_02\Algorithim_lab03_02\Debug\Algorith.1D5C2967.tlog\
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     文件         452  2019-04-15 15:03  Algorithim_lab03_02\Algorithim_lab03_02\源.cpp
     目录           0  2019-05-09 17:52  Algorithim_lab03_02\Debug\
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