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

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

一个简单的套模版的母函数题目，虽然很简单，但做这道题的时候很有意思，我写了好多遍，都是超时

直到在网上扒了个题解。。。。。

第一份代码是我写的，纯粹的套模版，

第二份代码是网上扒的，发现人家写的很是灵活，

我要反思反思，以后写题不能太死板了，要灵活处理。另外，能够灵活处理的前提是理解的足够深入，所以应该认真理解知识点

#include<bits/stdc++.h>
using namespace std;
const int MAXN=500000;
int num[6];
int a[MAXN],c[MAXN];
int solve()
{
    for(int i=0;i<=num[1];i++){
        a[i]=1;
        c[i]=0;
    }
    memset(c,0,sizeof(c));
    int i=0;
    while(1){
        if(i==2||i==5){
            for(int j=0;j<=MAXN;j++){
                for(int k=0;k<=num[i]*i;k+=i){
                    c[k+j]+=a[j];
                }
            }
            for(int j=0;j<=MAXN;j++){
                a[j]=c[j];
                c[j]=0;
            }
        }
        if(i==6) {break;}
        else i++;
    }
        for(int i=1;i<MAXN;i++){
            if(a[i]==0) return i;
        }
}
void print()
{
    for(int i=0;i<num[5]*5;i++){
        cout<<a[i]<<endl;
    }
}
int main()
{
    while(cin>>num[1]>>num[2]>>num[5]){
        if(num[1]==0&&num[2]==0&&num[5]==0){
            break;
        }
        cout<<solve()<<endl;
        //print();
        memset(num,0,sizeof(num));
    }
}

#include <cstdio>
#include <cstring>
#define M 8005

int vis[M];//标记可以组合的数；

int main()
{
    int num_1,num_2,num_5;
    while(true)
    {
        scanf("%d %d %d",&num_1,&num_2,&num_5);
        if(num_1 + num_2 + num_5 == 0) break;
        memset(vis,0,sizeof(vis));
        for(int i = 0; i <= num_1; i++)
            vis[i] = 1;
        for(int j = 0; j <= num_2 * 2; j += 2)
        {
            for(int k = 0; k <= num_1; k++)
            {
                vis[j + k] = 1;
            }
        }
        for(int j = 0; j <= num_5 * 5; j += 5)
        {
            for(int k = 0; k <= num_1 + num_2 * 2; k++)
            {
                if(vis[k]) //如果当前数可以组合，则加上；
                {
                    vis[k + j] = 1;
                }
            }
        }
        int i;
        for(i = 1; i <= num_1 + num_2 * 2 + num_5 * 5; i++)
        {
            if(vis[i] == 0)
            {
                break;
            }
        }
        printf("%d
",i);
    }
    return 0;
}