杭电acm 1085 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): 16934    Accepted Submission(s): 7620

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!
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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 30 0 0

Sample Output

4

这道题目也是用母函数，但也可以用其他方法做。由于这几天比较忙，具体的细节就不注释了，大家如果有什么问题可以留言，我每天都会登上去看一下，看到会回复的。

母函数：

#include <iostream>using namespace std;int num[8004],d[3004];int main(){    int A,B,C;    while(cin>>A>>B>>C){        if(A==0 && B==0 && C==0)            break;        int n=0;        memset(num,0,sizeof(num));        memset(d,0,sizeof(d));        for(int i=0;i<=A;i++)            for(int j=0;j<=B;j++)                d[i+j*2]=i+j*2;        for(int i=0;i<3004;i++)            for(int j=0;j<=C;j++)                num[d[i]+j*5]=d[i]+j*5;        int i;        for(i=1;i<8004;i++) if(num[i]==0) break;        cout<<i<<endl;    }    return 0;}

其他方法:

#include <iostream>using namespace std;int main(){    int A,B,C;    while(cin>>A>>B>>C){        if(A==0 && B==0 && C==0)            break;        if(A==0)            cout<<"1"<<endl;        else if(B<=1)             cout<<A+B*2+1<<endl;        else cout<<A+B*2+C*5+1<<endl;     }    return 0;}