HDU 1085 Holding Bin-Laden Captive!（思维，非母函数）

Holding Bin-Laden Captive!

题目链接

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23070 Accepted Submission(s): 10275

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

组合问题，典型的母函数模板题，奈何不会。
不过可以不用母函数写。
找规律

首先num_1=0时，无论后面的数怎样，结果肯定为1.
然后num_1+num_2*2小于4那么说求的值最大为4，直接num_1+num_2*2+1就行了。
最后就是满足上面的两个要求时，那结果就是比他们和大1的数了。num_1+num_2*2+num_5*5+1.

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
int main()
{
    int a,b,c;
    while(scanf("%d%d%d",&a,&b,&c)!=EOF)
    {
        if(a==0&&b==0&c==0) break;
        if(a==0) printf("1
");
        else if(a+2*b<4) printf("%d
",a+2*b+1);
        else printf("%d
",a+2*b+c*5+1);
    }
    return 0;
}