HDU 1085 Holding Bin-Laden Captive!(母函数，或者找规律）

Holding Bin-Laden Captive!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17653    Accepted Submission(s): 7902

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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一开始想用dp，但是dp[1000][1000][1000]开不了那么大（三维数组最大是[790][790][790]codeblocks亲测)，所以直接找规律。要细心，wa了两次，第一次是少了3,0,1这种情况，我的答案是9，正确答案是4。第二次是因为没有写&&n1+n2+n5>0 (orz...)

 

#include<queue>
#include<math.h>
#include<stdio.h>
#include<string.h>
#include<string>
#include<iostream>
#include<algorithm>
using namespace std;
#define N 1002
int n1,n2,n5;

int ziji(int n1,int n2,int n5)
{
    int ans;
    if(n1==0)ans=1;
    else if(n2==0&&n1<2)ans=2;
    else if(n1==1&&n2==1)ans=4;
    else if(n1==2&&n2==0&&n5>0)ans=3;
    else if(n1==3&&n2==0&&n5>0)ans=4;//第一次少了这种情况
    else
    {
        ans=n1+n2*2+n5*5+1;
    }
    return ans;
}

int main()
{
    while(~scanf("%d%d%d",&n1,&n2,&n5)&&n1+n2+n5>0)
    {
        cout<<ziji(n1,n2,n5)<<endl;
    }
    return 0;
}

母函数方法：