hdu1085 母函数的应用之每种硬币有个数限制

Holding Bin-Laden Captive!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8814    Accepted Submission(s): 3963

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
   
   
   
   
   
   
   
   

    
    
    
    

      题意： 
     
用1 2 5的硬币买不到的东西的价值（输出最小的）3者数量分别为num1 num2 num3
 
     
#include "stdio.h"
int c1[10000],c2[10000];
int main()
{
    int i,j,k,num1,num2,num3,n;
    while(scanf("%d %d %d",&num1,&num2,&num3)!=EOF)
    {
        if(!num1&&!num2&&!num3) break;
        n=num3*5+num2*2+num1;
        for(i=0;i<=n;i++)
        {
            c1[i]=0;
            c2[i]=0;
        }
        for (i=0;i<=num1;i++)
            c1[i]=1;
        for(j=0;j<=num1;j++)
   //    for (k=0;k+j<=num2*2;k=k+2)//num2*2表示最多可能达到的次数
            for (k=0;k<=num2*2;k=k+2)//num2*2表示最多可能达到的次数
            {  //注意  这里不要用k+j<=num2 因为以前求整数拆分输入数n 要保证次数小于n 因为最大为
    //x的n次方 表示为1个重为n的砝码     这里是求多项式相乘 不用限制了
                c2[j+k]+=c1[j];
            }
            for (j=0;j<=num2*2+num1;j++)//因为 下面的c2[j] 中的下标最大为j+k 由上一步可以看出 而j最大为num1 k最大为num2*2
            {                                          
                c1[j]=c2[j];                                         
                c2[j]=0;                         
            }
            for(j=0;j<=num2*2+num1;j++)//num1和num2*2不一定那个大那 要表示出所有的项数 所以要这样
                for(k=0;k<=num3*5;k+=5)
     c2[j+k]+=c1[j];
    for (j=0;j<=num3*5+num2*2+num1;j++)
    {                                          
     c1[j]=c2[j];                                         
     c2[j]=0;                         
    }
    for (i=0;i<=n;i++)
     if(!c1[i])  {printf("%d\n",i);break;}
     if(i==n+1) printf("%d\n",i);
    }
}