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 (硬幣) (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

 無疑是一道，母函式題.....（有關母函式的詳細，請檢視百度百科，在此就不做詳細的說明了）.

 這道題的方法為：g(x)=(1+x^1+x^2+......+x^num1)*x^2(1+x^1+....+x^num2)*x^5(1+x^1+.....+x^num3);

 其係數為0的即為所求：  所以關鍵是如何表達上面的式子.....一般採取的方法為先兩兩相乘...然後用得到結果與第三方相乘即可....

  程式碼如下：

 1 #include<iostream>
 2 #include<cstring>
 3 using namespace std;
 4 const int maxn=10001 ;
 5   int ratio[maxn],  //係數
 6     index[maxn];  //指數
 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("%dn",i);
43                   break;
44               }
45           }
46           if(judge)
47            printf("%dn",num1+2*num2+5*num3+1);
48       }
49      return 0;
50   }