[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 0x7f7f7ftypedef 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;}