POJ 3181 Dollar Dayz DP

f[i][j]=f[i-j][j]+f[i][j-1]，结果很大需要高精度。

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<sstream>
#include<cmath>
#include<climits>
#include<string>
#include<map>
#include<queue>
#include<vector>
#include<stack>
#include<set>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
#define pb(a) push(a)
#define INF 0x1f1f1f1f
#define lson idx<<1,l,mid
#define rson idx<<1|1,mid+1,r
#define PI  3.1415926535898
template<class T> T min(const T& a,const T& b,const T& c) {
    return min(min(a,b),min(a,c));
}
template<class T> T max(const T& a,const T& b,const T& c) {
    return max(max(a,b),max(a,c));
}
void debug() {
#ifdef ONLINE_JUDGE
#else

    freopen("in.txt","r",stdin);
    //freopen("d:\\out1.txt","w",stdout);
#endif
}
int getch() {
    int ch;
    while((ch=getchar())!=EOF) {
        if(ch!=' '&&ch!='\n')return ch;
    }
    return EOF;
}
const int MAX_LEN=20;
const int BASE=100000;
struct BigInt
{
    int bit[MAX_LEN];
    int len;
    BigInt(int n=0)
    {
        bit[0]=n;
        len=1;
    }
    BigInt operator + (const BigInt &ant)
    {
        int next=0;
        for(int i=0;i<len||i<ant.len;i++)
        {
            bit[i]=(i<len?bit[i]:0)+(i<ant.len?ant.bit[i]:0)+next;
            next=bit[i]/BASE;
            bit[i]%=BASE;
        }
        len=max(len,ant.len);
        if(next)bit[len++]=next;
        return *this;
    }
    void output()
    {
        printf("%d",bit[len-1]);
        for(int i=len-2;i>=0;i--)
            printf("%05d",bit[i]);
        printf("\n");
    }
};
BigInt dp[1005][105];
BigInt f(int n,int k)
{

    if(n==0)return BigInt(1);
    if(k==1)return BigInt(1);
    if(n<k)return f(n,n);
    if(dp[n][k].len!=1||dp[n][k].bit[0]!=0)return dp[n][k];
    dp[n][k]=f(n-k,k)+f(n,k-1);
    return dp[n][k];
}
int main()
{
    int n,k;
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        for(int i=0;i<=n;i++)
            for(int j=0;j<=k;j++)
                dp[i][j]=BigInt(0);
        f(n,k);
        dp[n][k].output();
    }
    return 0;
}