母函數--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): 25558    Accepted Submission(s): 11283

 

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

 

 

Recommend

We have carefully selected several similar problems for you:  1171 1398 2152 2082 1709 

 

題意：題目給出一些錢的數量，要求用給出的錢組合不能組成的最小錢數；
題解：直接套用母函數即可，把組合加法轉化成冪的加法；

#include<stdio.h>
#include<string.h>

int c[10000],temp[10000];
int cost[3]={1,2,5};
int num[3];

int main(){
	while(scanf("%d%d%d",&num[0],&num[1],&num[2])!=EOF){
		if(num[0]==0&&num[1]==0&&num[2]==0)	break;
        //初始化 
    	int total=num[0]*1+num[1]*2+num[2]*5;
    	memset(c,0,sizeof(c));
    	memset(temp,0,sizeof(temp));
    	for(int i=0;i<=num[0];i++){
    		c[i]=1; //-->(1+x+x^2+x^3+...+x^num[0])
    	} 
    	//核心代碼	
        for(int i=1;i<3;i++){
            for(int j=0;j<=total;j++){
                for(int k=0;k+j<=total&&k/cost[i]<=num[i];k+=cost[i])
				//k+j表示總金額數，它應該小於總和數
				//k/cost[i]表示k不能超過該種面值金額的數目 
                    temp[k+j]+=c[j];//模仿乘法分配率 
            }
            for(int j=0;j<=total;j++){
                c[j]=temp[j];
                temp[j]=0;
            }
        }
        //輸出 
    	for(int i=1;i<=total+1;i++)
    	if(!c[i]){
    	    printf("%d\n",i);
    	    break;
    	}
   }
   return 0;
}