HDUOJ-----1085Holding Bin-Laden Captive!

Holding Bin-Laden Captive!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 11977    Accepted Submission(s): 5354

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

Author

lcy

 无疑是一道，母函数题.....（有关母函数的详细，请查看百度百科，在此就不做详细的说明了）.

 这道题的方法为：g(x)=(1+x^1+x^2+......+x^num1)*x^2(1+x^1+....+x^num2)*x^5(1+x^1+.....+x^num3);

 其系数为0的即为所求：  所以关键是如何表达上面的式子.....一般采取的方法为先两两相乘...然后用得到结果与第三方相乘即可....

  代码如下：

 1 #include<iostream>
 2 #include<cstring>
 3 using namespace std;
 4 const int maxn=10001 ;
 5   int ratio[maxn],  //系数
 6     index[maxn];  //指数
 7 
 8   int main()
 9   {
10       int num1,num2,num3,i,j;
11       bool judge;
12       while(cin>>num1>>num2>>num3,num1+num2+num3)
13       {
14           memset(ratio,0,sizeof ratio);
15           memset(index,0,sizeof index);
16           for(i=0;i<=num1;i++)
17           {
18              ratio[i]=1;
19              for(j=0;j<=num2;j++)
20              {
21                  index[2*j+i]+=ratio[i];
22              }
23           }
24           for(i=0;i<=num1+2*num2;i++)
25           {
26               ratio[i]=index[i];
27               index[i]=0;
28           }
29           for(i=0;i<=num1+2*num2;i++)
30           {     
31               for(j=0;j<=num3;j++)
32               {
33                   index[j*5+i]=ratio[i];
34               }
35           }
36           judge=true;
37           for(i=0;i<=num1+2*num2+5*num3;i++)
38           {
39               if(!index[i])        
40               {
41                    judge=false;
42                   printf("%d\n",i);
43                   break;
44               }
45           }
46           if(judge)
47            printf("%d\n",num1+2*num2+5*num3+1);
48       }
49      return 0;
50   }