hdu-1085

http://acm.hdu.edu.cn/showproblem.php?pid=1085

Problem Description We all know that Bin-Laden is a notorious terrorist, and he has disappeared for a long time. But recently, it is reported that he hides in Hang Zhou of China! “Oh, God! How terrible! ”   Don’t be so afraid, guys. Although he hides in a cave of Hang Zhou, he dares not to go out. Laden is so bored recent years that he fling himself into some math problems, and he said that if anyone can solve his problem, he will give himself up! Ha-ha! Obviously, Laden is too proud of his intelligence! But, what is his problem? “Given some Chinese Coins (硬币) (three kinds-- 1, 2, 5), and their number is num_1, num_2 and num_5 respectively, please output the minimum value that you cannot pay with given coins.” You, super ACMer, should solve the problem easily, and don’t forget to take $25000000 from Bush!     Input Input contains multiple test cases. Each test case contains 3 positive integers num_1, num_2 and num_5 (0<=num_i<=1000). A test case containing 0 0 0 terminates the input and this test case is not to be processed.     Output Output the minimum positive value that one cannot pay with given coins, one line for one case.     Sample Input   1 1 3 0 0 0     Sample Output   4

 这个题的大致意思是说，刚开始给你面值为1 2 5三种硬币的个数，问你，用所给的硬币，不能组成的最小的硬币的面值是多少，这道题也是简单的组合问题，只不过和上一题不一样的是，以前给的物品是有无限多个的，而这次给定的物品的种类是有限的，所以这个要特别的注意一下，每个多项式的最大的项是多少，下面给出代码：

#include<stdio.h>
#include<string.h>
using namespace std;
const int maxn=10007;
int c1[maxn],c2[maxn];
int main()
{
    int a1,a2,a5;
    while(scanf("%d %d %d",&a1,&a2,&a5)&&(a1||a2||a5))
    {
    	int mm=a1+a2*2+a5*5;
        memset(c2,0,sizeof(c2));
        memset(c1,0,sizeof(c1));
        for(int i=0;i<=a1;i++)
            c1[i]=1;
        for(int i=2;i<=5;i++)
        {
            if(i==3||i==4) continue;
            for(int j=0;j<=mm;j++)
            {
                for(int k=0;k<=i*(i==2?a2:a5);k+=i)
                {
                    c2[j+k]+=c1[j];
                }
            }
            for(int j=0;j<=mm;j++)
                c1[j]=c2[j];
             memset(c2,0,sizeof(c2));
        }
        int q;
         for(int i=0;i<=mm;i++)
             {
                 if(c1[i]==0)
                 {
                     printf("%d\n",i);
                     break;
                 }
                 q=i;
             }
        if(q==mm) printf("%d\n",mm+1);
    }
}