option=com_onlinejudge&Itemid=8&category=471&page=show_problem&problem=4224" style="">题目链接:uva 1478 - Delta Wave

题目大意:对于每一个位置来说,能够向上,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:由于最后都要落回y=0的位置,所以上升的次数和下降的次数是同样的,而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∗i−2n)∗f(i−1)∗(n−2∗i+1)∗(n−2∗i+2)i∗(i+1)

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 (const char* number) {*this = number;} void DelZero ();
void Put (); void operator = (int number);
void operator = (char* number); bool operator < (const bign& b) const;
bool operator > (const bign& b) const { return b < *this; }
bool operator <= (const bign& b) const { return !(b < *this); }
bool operator >= (const bign& b) const { return !(*this < b); }
bool operator != (const bign& b) const { return b < *this || *this < b;}
bool operator == (const bign& b) const { return !(b != *this); } void operator ++ ();
void operator -- ();
bign operator + (const type& b);
bign operator + (const bign& b);
bign operator - (const type& b);
bign operator - (const bign& b);
bign operator * (const type& b);
bign operator * (const bign& b);
bign operator / (const type& b);
//bign operator / (const bign& b);
int operator % (const 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 - 2 * i + 1);
tmp = tmp / (1LL * i * (i + 1));;
ans = ans + tmp;
ans.len = min(ans.len, 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;
for (int i = len-1; i >= 0; i--) {
if (num[i] || flag) {
printf("%d", num[i]);
flag = true;
}
}
} void bign::operator = (char* number) {
len = strlen (number);
for (int i = 0; i < len; i++)
num[i] = number[len-i-1] - '0'; DelZero ();
} void bign::operator = (int number) { len = 0;
while (number) {
num[len++] = number%10;
number /= 10;
} DelZero ();
} bool bign::operator < (const bign& b) const {
if (len != b.len)
return len < b.len;
for (int i = len-1; i >= 0; i--)
if (num[i] != b.num[i])
return num[i] < b.num[i];
return false;
} void bign::operator ++ () {
int s = 1; for (int i = 0; i < len; i++) {
s = s + num[i];
num[i] = s % 10;
s /= 10;
if (!s) break;
} while (s) {
num[len++] = s%10;
s /= 10;
}
} void bign::operator -- () {
if (num[0] == 0 && len == 1) return; int s = -1;
for (int i = 0; i < len; i++) {
s = s + num[i];
num[i] = (s + 10) % 10;
if (s >= 0) break;
}
DelZero ();
} bign bign::operator + (const type& b) {
bign a = b;
return *this + a;
} bign bign::operator + (const bign& b) {
type bignSum = 0;
bign ans; for (int i = 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 % 10;
bignSum /= 10;
} return ans;
} bign bign::operator - (const type& b) {
bign a = b;
return *this - a;
} bign bign::operator - (const bign& b) {
type bignSub = 0;
bign ans;
for (int i = 0; i < len || i < b.len; i++) {
bignSub += num[i];
bignSub -= b.num[i];
ans.num[ans.len++] = (bignSub + 10) % 10;
if (bignSub < 0) bignSub = -1;
else bignSub = 0;
}
ans.DelZero ();
return ans;
} bign bign::operator * (const type& b) {
type bignSum = 0;
bign ans; ans.len = len;
for (int i = 0; i < len; i++) {
bignSum += num[i] * b;
ans.num[i] = bignSum % 10;
bignSum /= 10;
} while (bignSum) {
ans.num[ans.len++] = bignSum % 10;
bignSum /= 10;
}
return ans;
} bign bign::operator * (const bign& b) {
bign ans;
ans.len = 0; for (int i = 0; i < len; i++){
int bignSum = 0; for (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;
bignSum /= 10;
}
}
return ans;
} bign bign::operator / (const type& b) { bign ans; type s = 0;
for (int i = len-1; i >= 0; i--) {
s = s * 10 + num[i];
ans.num[i] = s/b;
s %= b;
} ans.len = len;
ans.DelZero ();
return ans;
} int bign::operator % (const int& b) { bign ans; int s = 0;
for (int i = len-1; i >= 0; i--) {
s = s * 10 + num[i];
ans.num[i] = s/b;
s %= b;
} return s;
}

uva 1478 - Delta Wave(递推+大数+卡特兰数+组合数学)的更多相关文章

  1. HDU 1023 Train Problem II (大数卡特兰数)

    Train Problem II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) ...

  2. Tiling(递推+大数)

    Description In how many ways can you tile a 2xn rectangle by 2x1 or 2x2 tiles? Here is a sample tili ...

  3. Children’s Queue HDU 1297 递推+大数

    题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1297 题目大意: 有n个同学, 站成一排, 要求 女生最少是两个站在一起, 问有多少种排列方式. 题 ...

  4. UVa 12034 - Race(递推 + 杨辉三角)

    链接: https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem& ...

  5. ACM学习历程—HDU1041 Computer Transformation(递推 && 大数)

    Description A sequence consisting of one digit, the number 1 is initially written into a computer. A ...

  6. UVA 10288 - Coupons(概率递推)

    UVA 10288 - Coupons option=com_onlinejudge&Itemid=8&page=show_problem&category=482&p ...

  7. uva 11375 Matches (递推)

    http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&p ...

  8. UVa 10561 (SG函数 递推) Treblecross

    如果已经有三个相邻的X,则先手已经输了. 如果有两个相邻的X或者两个X相隔一个.,那么先手一定胜. 除去上面两种情况,每个X周围两个格子不能再放X了,因为放完之后,对手下一轮再放一个就输了. 最后当“ ...

  9. UVA 557 - Burger(概率 递推)

     Burger  When Mr. and Mrs. Clinton's twin sons Ben and Bill had their tenth birthday, the party was ...

随机推荐

  1. 最小公倍数 【杭电-HDOJ-1108】 附题+具体解释

    /* 最小公倍数 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total S ...

  2. js取消radio选中 反选

    var radio=document.createElement("input");radio.type="radio";radio.onclick = fun ...

  3. JavaScript Map 实现

    //定义map function Map() { this.container = {}; } //将key-value放入map中 Map.prototype.put = function(key, ...

  4. flashback database(drop tablespace)

    1.首先记录时间 select to_char(systimestamp,'yyyy-mm-dd HH24:MI:SS') from dual;--2014-04-25 13:55:48 查看表sel ...

  5. java基础讲解09-----接口,继承,多态

    还有什么包装类,数字类,这些简单的我就不想过去介绍,前面也大概的介绍了下,继承,多态 1.类的继承 继承的思想:基于某个父类的扩展,制定一个新的子类.子类可以继承父类原有的属性,方法,也可以重写父类的 ...

  6. 关闭MongoDB

    以下方法干净地关闭MongoDB: 完成所有挂起的操作.刷新数据到数据文件.关闭所有的数据文件 1. > use admin switched to db admin > db.shutd ...

  7. c++ 返回对象的引用要小心

    除非能保证返回对象的生命周期足够长. 一定不要返回临时对象的引用.

  8. Atitit.ati  str  字符串增强api

    Atitit.ati  str  字符串增强api 1. java StringUtils方法全览 分类: Java2011-11-30 17:22 8194人阅读 评论(2) 收藏 举报 javas ...

  9. Android 开发手记二 C可执行程序编译实例(转帖)

    http://www.cnblogs.com/gaozehua/archive/2011/09/02/2164077.html

  10. 蓝牙(CoreBluetooth)-概述

    蓝牙(CoreBluetooth)-概述 通过此框架可以让你的Mac和iOS应用程序与外部蓝牙设备通信 外部设备: 就是需要通过iOS App控制器的其他设备: 例如:心率检测仪.数字温控器 蓝牙通讯 ...