HDOJ 1085 母函数

Holding Bin-Laden Captive!

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23675 Accepted Submission(s): 10544

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

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题目大概意思是有1 2 5硬币num1 num2 num5个。问最小不能组合的数位多少。
就是问母函数总系数为0的最低幂是哪项。
这个题目就对模板有了大量的修改。主要在于范围的改动，已经将最外面本来选择项的循环直接拆开（因为1 2 5 不好写）如果不拆开可以用数组也是没问题的。

#include
#define max(a,b) a>b?a:b
#define min(a,b) a
using namespace std ;
typedef long long ll;
typedef unsigned long long ull;

const int MAX = 10000+5; 

int c1[MAX],c2[MAX];
int main(){ 
    int num1,num2,num5;
    while(cin>>num1>>num2>>num5){
        if(!num1&&!num2&&!num5) break;
        for(int i = 0 ; i <= num1*1+num2*2+num5*5 ; i++){
            c1[i] = 0;
            c2[i] = 0;
        }
        for(int i = 0 ; i <= num1*1 ; i++)
            c1[i] = 1;
        for(int i = 0 ; i <= num1*1 ; i++)
            for(int j = 0 ; j <= num2*2 ; j+=2){
                c2[j+i] += c1[i];
            }
        for(int i = 0 ; i <= num1*1+num2*2 ; i++){
            c1[i] = c2[i];
            c2[i] = 0;
        }
        for(int i = 0 ; i <= num1*1+num2*2 ; i++)
            for(int j = 0 ; j <= num5*5 ; j+=5){
                c2[i+j] += c1[i];
            }
        for(int i = 0 ; i <= num1*1+num2*2+num5*5 ; i++){
            c1[i] = c2[i];
            c2[i] = 0;
        }
        int i ; 
        for(i=0; i<=num1*1+num2*2+num5*5 ; i++)
            if(c1[i] == 0){
                cout<break;
            }
        if(i == num1*1+num2*2+num5*5+1)
            cout<return 0;
}