uva 1478 - Delta Wave(递推+大数+卡特兰数+组合数学)

题目链接：uva 1478 - Delta Wave

题目大意：对于每个位置来说，可以向上，水平，向下，坐标不能位负，每次上下移动最多为1， 给定n问说有多少种不同的图。结果对10100取模。

解题思路：因为最后都要落回y=0的位置，所以上升的次数和下降的次数是相同的，并且上升下降的关系满足出栈入栈的关系。即卡特兰数。
所以每次枚举i，表示有i个上升，i个下降，用组合数学枚举出位置，然后累加求和。

C(2∗in)∗f(i)=C(−2−1∗n2i+∗())
注意取模后的前导0

#include <cstdio>
#include <cstring>
#include <iostream>

using namespace std;
typedef long long  type;
const int MAXN = 10005;

struct bign {
    int len,num[MAXN];

    bign () {
        len = 0;
        memset(num,0,sizeof(num));
    }
    bign (int number) {*this = number;}
    bign (char* number) {*this = number;}

    void DelZero ();
    void Put ();

    void operator = (int number);
    char* number);

    bool operator <  (const bign& b) const;
    operator >  (const { return b < *this; }
    operator <= (return !(b < *this); }
    operator >= (return !(*this < b); }
    operator != (this || *this < b;}
    operator == (return !(b != *this); }

    operator ++ ();
    operator -- ();
    bign operator + (const type& b);
    bign const bign& b);
    bign operator - (operator * (operator / (const type& b);
    //bign operator / (const bign& b);
    int operator % (int& b);
};

/*Code*/
int main () {
    int n;
    while (scanf("%d",&n) == 1) {
        bign ans = 0;
        bign tmp = 1;
        ans = ans + tmp;
        for (int i = 1; i <= n/2; i++) {
            tmp = tmp * 1LL * (n - 2 * i + 2) * (n - 1);
            tmp = tmp / (1LL * i * (i + 1));;
            ans = ans + tmp;
            ans.len = min(ans.len,255)">100);
        }
        ans.Put();
        printf("\n");
    }
}

void bign::DelZero () {
    while (len && num[len-1] == 0)
        len--;

    if (len == 0)
        num[len++] = 0;
}

void bign::Put () {
    bool flag = false;
    int i = len-1; i >= 0; i--) {
        if (num[i] || flag) {
            true;
        }
    }
}

void bign::char* number) {
    len = strlen (number);
    0; i < len; i++)
        num[i] = number[len-i-1] - '0';

    DelZero ();
}

int number) {

    len = 0;
    while (number) {
        num[len++] = number%10;
        number /= 10;
    }

    DelZero ();
}

bool bign::operator < (const {
    if (len != b.len)
        return len < b.len;
    0; i--)
        if (num[i] != b.num[i])
            return num[i] < b.num[i];
    return false;
}

operator ++ () {
    int s = 1;

    0; i < len; i++) {
        s = s + num[i];
        num[i] = s % 10;
        s /= 10;
        if (!s) break;
    }

    while (s) {
        num[len++] = s%10;
    }
}

operator -- () {
    if (num[0] == 0 && len == 1) return;

    int s = -1;
    0; i < len; i++) {
        s = s + num[i];
        num[i] = (s + 10) % if (s >= 0) break;
    }
    DelZero ();
}

bign bign::const type& b) {
    bign a = b;
    return *this + a;
}

bign bign::const bign& b) {
    type bignSum = 0;
    bign ans;

    0; i < len || i < b.len; i++) {
        if (i < len) bignSum += num[i];
        if (i < b.len) bignSum += b.num[i];

        ans.num[ans.len++] = bignSum % 10;
        bignSum /= 10;
    }

    while (bignSum) {
        ans.num[ans.len++] = bignSum % return ans;
}

bign bign::this - a;
}


bign bign::const bign& b) {
    type bignSub = 0;
    bign ans;
    0; i < len || i < b.len; i++) {
        bignSub += num[i];
        bignSub -= b.num[i];
        ans.num[ans.len++] = (bignSub + if (bignSub < 0) bignSub = -1;
        else bignSub = 0;
    }
    ans.DelZero ();
    const type& b) {
    type bignSum = 0;
    bign ans;

    ans.len = len;
    0; i < len; i++) {
        bignSum += num[i] * b;
        ans.num[i] = bignSum % 10;
    }
    const bign& b) {
    bign ans;
    ans.len = 0; 

    0; i < len; i++){  
        int bignSum = 0;  

        int j = 0; j < b.len; j++){  
            bignSum += num[i] * b.num[j] + ans.num[i+j];  
            ans.num[i+j] = bignSum % 10;  
            bignSum /= 10;
        }  
        ans.len = i + b.len;  

        while (bignSum){  
            ans.num[ans.len++] = bignSum % 10;
        }  
    }  
    const type& b) {

    bign ans;

    type s = 0; i--) {
        s = s * 10 + num[i];
        ans.num[i] = s/b;
        s %= b;
    }

    ans.len = len;
    ans.DelZero ();
    return ans;
}

int bign::int& b) {

    bign ans;

    10 + num[i];
        ans.num[i] = s/b;
        s %= b;
    }

    return s;
}