uva 1478 - Delta Wave(递推+大数+卡特兰数+组合数学)
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
- 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) ...
- 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 ...
- 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 ...
- 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& ...
- 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 ...
- UVA 10288 - Coupons(概率递推)
UVA 10288 - Coupons option=com_onlinejudge&Itemid=8&page=show_problem&category=482&p ...
- uva 11375 Matches (递推)
http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&p ...
- 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“ ...
- 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
- tar 命令详解 / xz 命令
]# tar [-cxtzjvfpPN] 文件与目录 ....参数:-c :建立一个压缩文件的参数指令(create 的意思):-x :解开一个压缩文件的参数指令!-t :查看 tarfile 里面的 ...
- 实现对数据进行分组小计并计算合计的实例 asp.net
可以通过数据绑定来实现 通过union all 来实现数据库 SELECT * FROM v3_pay_list2 where ( (ought_date >= '2012-12-06') a ...
- iperf使用
1. sourceforge搜索iperf下载 2. ./configure make make install 3. server:iperf -s -p 12345 -i 1 -M: client ...
- javascript中call apply的区别
obj.call(thisObj, arg1, arg2, ...); obj.apply(thisObj, [arg1, arg2, ...]); 两者作用一致,都是把obj(即this)绑定到th ...
- JUC组件扩展(二)-JAVA并行框架Fork/Join(二):同步和异步
在Fork/Join框架中,提交任务的时候,有同步和异步两种方式. invokeAll()的方法是同步的,也就是任务提交后,这个方法不会返回直到所有的任务都处理完了. fork方法是异步的.也就是你提 ...
- CentOS安装自动补全安装包
CentOS7标准版有这个功能,但是CentOS6却没有,其实很简单: 1.安装bash-completion yum install bash-completion 2.保存一下最新的缓存 yum ...
- web ul li
<html> <head> <style type="text/css"> ul{float:right} ul li{float:left; ...
- oracle中查看正在运行的并行进程
select count(*) from v$px_process a where a.STATUS='IN USE';
- Javaweb开发中关于不同地方出现的绝对路径和相对路径
1.转发和包含路径 a)以“/”开头:相对当前项目路径,即默认为http://localhost:8080/项目名/ b)不以“/”开头:相对当前Servlet路径. eg:在Aservlet中写“B ...
- Jquery学习笔记(1)--JQuery原理,与JS对象互换,核心函数
js对象转jQuery对象,$('num'), jQuery对象转js对象,$('num')[0],或$('num').get(0). 1.点击换行,each(),html(),attr(),每个h1 ...