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  • HDU1085(母函数)

    大意: 面值为1分,2分,3分的硬币各有a,b,c枚,求不能用这些硬币表示的最小值。

    分析:硬币能够表示的最大值max=1*a+2*b+5*c,计算1,2,3...max,max+1的系数是否为0,若0则不能表示。

    代码

    #include <iostream>
    #include <cstdio>
    #include <algorithm>
    #include <cstring>
    #include <cmath>
    using namespace std;
    int c1[10050], c2[10050];
    int main()
    {
    	int a, b, c;
    	while (~scanf("%d%d%d", &a, &b, &c))
    	{
    		if (a == 0 && b == 0 && c == 0)
    			break;
    		memset(c1, 0, sizeof(c1));
    		memset(c2, 0, sizeof(c2));
    		for (int i = 0; i <= a; i++)
    			c1[i] = 1;
    		for (int i = 0; i <= a; i++)
    			for (int j = 0; j <= b; j++)
    				c2[i + 2 * j] += c1[i];
    		for (int i = 0; i <= (a + 2 * b); i++)
    		{
    			c1[i] = c2[i];
    			c2[i] = 0;
    		}
    		for (int i = 0; i <= (a + 2 * b); i++)
    			for (int j = 0; j <= c; j++)
    				c2[i + 5 * j] += c1[i];
    		for (int i = 0; i <= (a + 2 * b + 5 * c); i++)
    			c1[i] = c2[i];
    		for (int i = 0; i <= (a + 2 * b + 5 * c+1); i++)
    			if (c1[i] == 0)
    			{
    				printf("%d
    ",i);
    				break;
    			}
    	}
    	return 0;
    }

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  • 原文地址:https://www.cnblogs.com/nickqiao/p/7583377.html
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