hdu 1085 Holding Bin-Laden Captive! (母函数)
http://acm.hdu.edu.cn/showproblem.php?pid=1085 题意;一元的钱有num_1张,2元的钱有num_2张,5元的钱有num_5张,问最小的不能组成的钱是多少。 思路:有限个个数的母函数,并且不知道最好要多少,所以限制条件变成了不同种类钱的个数。统计0到num_1+2num_2+5num_5的方案数,第一个为0的就是答案。
20161117更新:之前贴的代码好像有点问题...估计是最后一次更新以后忘记保存了orz
/* ***********************************************
Author :111qqz
Created Time :2016年02月25日 星期四 22时26分16秒
File Name :code/hdu.1085.cpp
************************************************ */
1#include <cstdio>
2#include <cstring>
3#include <iostream>
4#include <algorithm>
5#include <vector>
6#include <queue>
7#include <set>
8#include <map>
9#include <string>
10#include <cmath>
11#include <cstdlib>
12#include <ctime>
13#define fst first
14#define sec second
15#define lson l,m,rt<<1
16#define rson m+1,r,rt<<1|1
17#define ms(a,x) memset(a,x,sizeof(a))
18typedef long long LL;
19#define pi pair < int ,int >
20#define MP make_pair
1using namespace std;
2const double eps = 1E-8;
3const int dx4[4]={1,0,0,-1};
4const int dy4[4]={0,-1,1,0};
5const int inf = 0x3f3f3f3f;
6const int N=9E3+7;
7int num_1,num_2,num_5;
8int a[N],tmp[N];
9int main()
10{
11 #ifndef ONLINE_JUDGE
12 freopen("code/in.txt","r",stdin);
13 #endif
1 while (~scanf("%d%d%d",&num_1,&num_2,&num_5))
2 {
3 if (num_1==0&&num_2==0&&num_5==0) break;
4 ms(a,0);
5 int total = num_1+2*num_2+5*num_5;
6 for ( int i = 0 ; i <= num_1; i++)
7 {
8 a[i] = 1;
9 tmp[i] = 0;
10 }
1 for ( int j = 0 ; j <= num_1 ; j++)
2 {
3 for ( int k = 0; k <=num_2 ; k++)
4 {
5 tmp[j+2*k] += a[j];
6 }
7 }
1 for ( int j = 0 ; j <= num_1 +2*num_2 ; j++)
2 {
3 a[j] = tmp[j];
4 tmp[j] = 0 ;
5 }
1 for ( int j = 0 ; j <= num_1+2*num_2 ; j++)
2 {
3 for ( int k = 0; k <= num_5 ; k++)
4 {
5 tmp[j+5*k]+=a[j];
6 }
7 }
8 for ( int j = 0 ; j <= total ; j++)
9 {
10 a[j] = tmp[j];
11 tmp[j] = 0 ;
12 }
1 for ( int i = 0 ; i <= total +1 ; i ++)
2 {
3 if (a[i]==0)
4 {
5 printf("%d\n",i);
6 break;
7 }
8 }
9 }
1 #ifndef ONLINE_JUDGE
2 fclose(stdin);
3 #endif
4 return 0;
5}