BZOJ 1655: [Usaco2006 Jan] Dollar Dayz 奶牛商店 (母函数，高精度)
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。。。
/* ***********************************************
Author :111qqz
Created Time :2016年04月14日 星期四 16时42分31秒
File Name :code/bzoj/1655.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=1E4+7;
7const int MAXN =300;
8int n,m;
9int a[N][105],tmp[N][105];
1void add(int x,int y)
2{
3 for ( int i = 1 ; i <= 100 ; i++)
4 tmp[x][i]+=a[y][i];
1 for ( int i = 1 ; i <= 100 ; i++)
2 {
3 tmp[x][i+1]+=tmp[x][i]/10;
4 tmp[x][i]%=10;
5 }
}
1void tran( int x)
2{
3 for ( int i = 1 ; i <= 100 ; i++)
4 {
5 a[x][i] = tmp[x][i];
6 tmp[x][i] = 0;
7 }
8}
1int main()
2{
3 #ifndef ONLINE_JUDGE
4 freopen("code/in.txt","r",stdin);
5 #endif
1 scanf("%d %d",&n,&m);
2 ms(a,0);
3 m = min(m,n);
4 for ( int i = 0 ; i <= n ; i++)
5 {
6 a[i][1] = 1;
7 tmp[i][1] = 0 ;
8 }
1 for ( int i = 2 ; i <= m ; i++)
2 {
3 for ( int j = 0 ;j <= n ; j++)
4 {
5 for ( int k = 0 ; k+j<= n ; k+=i)
6 {
7 for ( int z = 1 ; z <= 40 ; z++)
8 {
9 tmp[k+j][z] += a[j][z];
10 }
1 }
2 }
3// cout<<"aaa"<<endl;
1 for ( int j = 0 ; j <= n ; j++)
2 {
3 for ( int z = 1 ; z <= 40 ; z++)
4 {
5 tmp[j][z+1]+=tmp[j][z]/10;
6 tmp[j][z]%=10;
7 }
8 }
9 for ( int j = 0 ; j <= n ; j++)
10 {
11 for ( int z = 1 ; z <= 40 ; z++)
12 {
13 a[j][z] = tmp[j][z];
14 tmp[j][z] = 0;
15 }
16 }
17 }
1 int len = 100;
2 while (a[n][len]==0) len--;
3// cout<<"len:"<<len<<endl;
4 for ( int i = len ;i >0 ; i--) printf("%d",a[n][i]);
1 #ifndef ONLINE_JUDGE
2 fclose(stdin);
3 #endif
4 return 0;
5}