1655: [Usaco2006 Jan] Dollar Dayz 奶牛商店#
Time Limit: 5 Sec Memory Limit: 64 MB Submit: 353 Solved: 190 [Submit][Status][Discuss]
Description#
Farmer John goes to Dollar Days at The Cow Store and discovers an unlimited number of tools on sale. During his first visit, the tools are selling variously for $1, $2, and $3. Farmer John has exactly $5 to spend. He can buy 5 tools at $1 each or 1 tool at $3 and an additional 1 tool at $2. Of course, there are other combinations for a total of 5 different ways FJ can spend all his money on tools. Here they are: 1 @ US$3 + 1 @ US$2 1 @ US$3 + 2 @ US$1 1 @ US$2 + 3 @ US$1 2 @ US$2 + 1 @ US$1 5 @ US$1 Write a program than will compute the number of ways FJ can spend N dollars (1 <= N <= 1000) at The Cow Store for tools on sale with a cost of $1..$K (1 <= K <= 100).
约翰到奶牛商场里买工具．商场里有K(1≤K≤100).种工具，价格分别为1，2，…，K美元．约翰手里有N(1≤N≤1000)美元，必须花完．那他有多少种购买的组合呢？
Input#
A single line with two space-separated integers: N and K.
仅一行，输入N，K.
Output#
A single line with a single integer that is the number of unique ways FJ can spend his money.
不同的购买组合数．
Sample Input#
5 3
Sample Output#
5
思路：母函数裸题，还卡个高精度。。差评。。。
代码实现
1/* ***********************************************
2Author :111qqz
3Created Time :2016年04月14日 星期四 16时42分31秒
4File Name :code/bzoj/1655.cpp
5************************************************ */
6
7#include <cstdio>
8#include <cstring>
9#include <iostream>
10#include <algorithm>
11#include <vector>
12#include <queue>
13#include <set>
14#include <map>
15#include <string>
16#include <cmath>
17#include <cstdlib>
18#include <ctime>
19#define fst first
20#define sec second
21#define lson l,m,rt<<1
22#define rson m+1,r,rt<<1|1
23#define ms(a,x) memset(a,x,sizeof(a))
24typedef long long LL;
25#define pi pair < int ,int >
26#define MP make_pair
27
28using namespace std;
29const double eps = 1E-8;
30const int dx4[4]={1,0,0,-1};
31const int dy4[4]={0,-1,1,0};
32const int inf = 0x3f3f3f3f;
33const int N=1E4+7;
34const int MAXN =300;
35int n,m;
36int a[N][105],tmp[N][105];
37
38void add(int x,int y)
39{
40 for ( int i = 1 ; i <= 100 ; i++)
41 tmp[x][i]+=a[y][i];
42
43 for ( int i = 1 ; i <= 100 ; i++)
44 {
45 tmp[x][i+1]+=tmp[x][i]/10;
46 tmp[x][i]%=10;
47 }
48
49}
50
51void tran( int x)
52{
53 for ( int i = 1 ; i <= 100 ; i++)
54 {
55 a[x][i] = tmp[x][i];
56 tmp[x][i] = 0;
57 }
58}
59
60int main()
61{
62 #ifndef ONLINE_JUDGE
63 freopen("code/in.txt","r",stdin);
64 #endif
65
66 scanf("%d %d",&n,&m);
67 ms(a,0);
68 m = min(m,n);
69 for ( int i = 0 ; i <= n ; i++)
70 {
71 a[i][1] = 1;
72 tmp[i][1] = 0 ;
73 }
74
75 for ( int i = 2 ; i <= m ; i++)
76 {
77 for ( int j = 0 ;j <= n ; j++)
78 {
79 for ( int k = 0 ; k+j<= n ; k+=i)
80 {
81 for ( int z = 1 ; z <= 40 ; z++)
82 {
83 tmp[k+j][z] += a[j][z];
84 }
85
86 }
87 }
88// cout<<"aaa"<<endl;
89
90 for ( int j = 0 ; j <= n ; j++)
91 {
92 for ( int z = 1 ; z <= 40 ; z++)
93 {
94 tmp[j][z+1]+=tmp[j][z]/10;
95 tmp[j][z]%=10;
96 }
97 }
98 for ( int j = 0 ; j <= n ; j++)
99 {
100 for ( int z = 1 ; z <= 40 ; z++)
101 {
102 a[j][z] = tmp[j][z];
103 tmp[j][z] = 0;
104 }
105 }
106 }
107
108 int len = 100;
109 while (a[n][len]==0) len--;
110// cout<<"len:"<<len<<endl;
111 for ( int i = len ;i >0 ; i--) printf("%d",a[n][i]);
112
113
114 #ifndef ONLINE_JUDGE
115 fclose(stdin);
116 #endif
117 return 0;
118}