[Usaco2006 Jan] Dollar Dayz 奶牛商店

Description

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

---

一个简单的dp，不过要写高精度

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x7f7f7f
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
using namespace std;
inline int read(){
	int x=0,f=1;char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar())  if (ch=='-')	f=-1;
	for (;ch>='0'&&ch<='9';ch=getchar())   x=(x<<3)+(x<<1)+ch-'0';
	return x*f;
}
inline void print(int x){
	if (x>=10) print(x/10);
	putchar(x%10+'0');
}
const int N=1e4;
const int base=1e6;
const int digit=6;
const int maxn=1e2;
char s[maxn*digit];
struct Bignum{
	int len,v[maxn];
	void read(){
		scanf("%s",s);
		memset(v,0,sizeof(v));
		int n=strlen(s),tim=1;
		len=(n-1)/digit+1;
		for (int i=0,j=n-1;i<j;i++,j--)  swap(s[i],s[j]);
		for (int i=0;i<n;i++){
			v[i/digit]+=(s[i]-'0')*tim,tim*=10;
			if (tim==base)  tim=1;
		}
	}
	void write(){
		printf("%d",v[len-1]);
		for (int i=len-2;~i;i--)	printf("%0*d",digit,v[i]);
		putchar('\n');
	}
}f[N+10];
Bignum operator +(const Bignum &x,const Bignum &y){
	Bignum z;
	memset(z.v,0,sizeof(z.v));
	z.len=max(x.len,y.len);
	for (int i=0;i<=z.len;i++)   z.v[i]+=x.v[i]+y.v[i],z.v[i+1]=z.v[i]/base,z.v[i]%=base;
	while (z.v[z.len])  z.v[z.len+1]=z.v[z.len]/base,z.v[z.len]%=base,z.len++;
	return z;
}
int main(){
	int n=read(),k=read();
	f[0].len=f[0].v[0]=1;
	for (int i=1;i<=k;i++)
		for (int j=i;j<=n;j++)
			f[j]=f[j]+f[j-i];
	f[n].write();
	return 0;
}