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('
');
}
}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;
}