wenbao与卡特兰数

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LL h[maxn];

LL exp_mod(LL a){
    LL b = MOD - 2;
    LL res = 1;
    while (b){
        if (b & 1)
            res = (res*a) % MOD;
        b >>= 1;
        a = (a*a) % MOD;
    }
    return res;
}

void init(){
    LL i;
    h[0] = h[1] = 1;
    for (i = 2; i < maxn; i++)
        h[i] = (((4 * i - 2) % MOD * exp_mod(i + 1)) % MOD * h[i - 1]) % MOD;
}

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http://acm.hdu.edu.cn/showproblem.php?pid=1023

卡特兰数，非常好的一个练习大数的题目

说到大数当然首推java（python忽略）

import java.util.*;
import java.math.*;
public class Main{
    private static BigInteger a[] = new BigInteger [105];
    public static void init(){
        BigInteger x = new BigInteger ("1");
        a[0] = x;
        for(int i = 1; i <= 100; ++i){
            Integer xxx = 4*i-2;
            Integer y = i+1;
            BigInteger xx = new BigInteger(xxx.toString()), yy = new BigInteger(y.toString());
            a[i] = (a[i-1].multiply(xx)).divide(yy);
            //System.out.println(a[i]);
        }
    }
    public static void main(String args[]){
        init();
        Scanner cin = new Scanner(System.in);
        while(cin.hasNext()){
            int b = cin.nextInt();
            System.out.println(a[b]);
        }
    }
}

不得不说java的包真的厉害，毕竟是大牛写的，优化到极致。

所以说要多看源码，多向大牛看齐！！！

我们可以将一个大数转化成为一个尽量大的n进制数（充分利用内存），将每一位放在数组里面。OK！就这么简单

#include <iostream>
#include <string.h>
using namespace std;
const int maxn = 105;
const int Mod = 1e5;
int f[maxn][maxn];
void mul(int a[], int x){
    int sum = 0;
    for(int i = maxn-1; i >= 0; --i){
        sum += a[i]*x;
        a[i] = sum%Mod;
        sum /= Mod;
    }
}
void div(int a[], int x){
    int sum = 0;
    for(int i = 0; i < maxn; ++i){
        sum = (sum*Mod + a[i]);
        a[i] = sum/x;
        sum %= x;
    }
}
void init(){
    memset(f[1], 0, sizeof(int)*maxn);
    f[1][maxn-1] = 1;
    for(int i = 2; i <= 100; ++i){
        memcpy(f[i], f[i-1], sizeof(int)*maxn);
        mul(f[i], 4*i-2);
        div(f[i], i+1);
    }
}
void pri(int x){
    int i;
    for(i = 0; !f[x][i]; ++i);
    printf("%d", f[x][i]);
    i++;
    for(; i < maxn; ++i){
        printf("%05d", f[x][i]);
    }
    printf("
");
}
int main(){
    init();
    int x;
    while(~scanf("%d", &x)){
        pri(x);
    }
    return 0;
}

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只有不断学习才能进步！