hdu 1085 给出数量限制的母函数问题 Holding Bin-Laden Captive!

Holding Bin-Laden Captive!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17256    Accepted Submission(s): 7734

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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#include<stdio.h>
#include<string.h>
int c[],temp[];
int cost[]={,,};
int num[];
int main(){
   while(scanf("%d%d%d",&num[],&num[],&num[])!=EOF){
      if(num[]==&&num[]==&&num[]==)
        break;
        int total=num[]*+num[]*+num[]*;
      memset(c,,sizeof(c));
      memset(temp,,sizeof(temp));
      for(int i=;i<=num[];i++)
        c[i]=;

        for(int i=;i<;i++){
            for(int j=;j<=total;j++){
                for(int k=;k+j<=total&&k/cost[i]<=num[i];k+=cost[i])///此步应该特别注意，要保证k/cost[i]〈num[i]   即k的总值不能超过题里给出的范围
                    temp[k+j]+=c[j];
            }

            for(int ii=;ii<=total;ii++){
                c[ii]=temp[ii];
                temp[ii]=;
            }
        }
    for(int i=;i<=total+;i++)
    if(!c[i]){
        printf("%d\n",i);
        break;
    }
   }
   return ;
}