标签：enter idt max lin auth pac not proc void
Buy the Ticket |
| Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) |
| Total Submission(s): 1886 Accepted Submission(s): 832 |
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Problem Description
The \\\\\\\"Harry Potter and the Goblet of Fire\\\\\\\" will be on show in the next few days. As a crazy fan of Harry Potter, you will go to the cinema and have the first sight, won’t you?
Suppose the cinema only has one ticket-office and the price for per-ticket is 50 dollars. The queue for buying the tickets is consisted of m + n persons (m persons each only has the 50-dollar bill and n persons each only has the 100-dollar bill). Now the problem for you is to calculate the number of different ways of the queue that the buying process won\\\\\\\‘t be stopped from the first person till the last person. Note: initially the ticket-office has no money. The buying process will be stopped on the occasion that the ticket-office has no 50-dollar bill but the first person of the queue only has the 100-dollar bill. |
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Input
The input file contains several test cases. Each test case is made up of two integer numbers: m and n. It is terminated by m = n = 0. Otherwise, m, n <=100.
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Output
For each test case, first print the test number (counting from 1) in one line, then output the number of different ways in another line. |
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Sample Input
3 0 3 1 3 3 0 0 |
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Sample Output
Test #1: 6 Test #2: 18 Test #3: 180 |
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Author
HUANG, Ninghai
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Recommend
Eddy
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这个题！！！我真是*了狗了，推了一晚上结果推出来一看是卡特兰数，用大数写完还各种漏洞。。。。。。
/* 由以上两步能得出来规律先不管买票的顺序：在n×m的矩阵中dp[i][j]=dp[i-1][j]+dp[i][j-1]; 然后你就会惊奇的发现这个竟然也跟卡特兰数有关，真**厉害，得到：C(m+n)(m)-C(m+n)(m+1); 然后再考虑买票的顺序再乘上n！*m！，化简完了就是(m+n)! * (m-n+1)/(m+1) */ #include<bits/stdc++.h> using namespace std; /* * 完全大数模板 * 输出cin>>a * 输出a.print(); * 注意这个输入不能自动去掉前导0的，可以先读入到char数组，去掉前导0，再用构造函数。 */ #define MAXN 9999 #define MAXSIZE 1010 #define DLEN 4 class BigNum { private: int a[500]; //可以控制大数的位数 int len; public: BigNum() { len=1;memset(a,0,sizeof(a)); } //构造函数 BigNum(const int); //将一个int类型的变量转化成大数 BigNum(const char*); //将一个字符串类型的变量转化为大数 BigNum(const BigNum &); //拷贝构造函数 BigNum &operator=(const BigNum &); //重载赋值运算符，大数之间进行赋值运算 friend istream& operator>>(istream&,BigNum&); //重载输入运算符 friend ostream& operator<<(ostream&,BigNum&); //重载输出运算符 BigNum operator+(const BigNum &)const; //重载加法运算符，两个大数之间的相加运算 BigNum operator-(const BigNum &)const; //重载减法运算符，两个大数之间的相减运算 BigNum operator*(const BigNum &)const; //重载乘法运算符，两个大数之间的相乘运算 BigNum operator/(const int &)const; //重载除法运算符，大数对一个整数进行相除 运算 BigNum operator^(const int &)const; //大数的n次方运算 int operator%(const int &)const; //大数对一个int类型的变量进行取模运算 bool operator>(const BigNum &T)const; //大数和另一个大数的大小比较 bool operator>(const int &t)const; //大数和一个int类型的变量的大小比较 void print(); //输出大数 }; BigNum::BigNum(const int b) //将一个int类型的变量转化为大数 { int c,d=b; len=0; memset(a,0,sizeof(a)); while(d>MAXN) { c=d-(d/(MAXN+1))*(MAXN+1); d=d/(MAXN+1); a[len++]=c; } a[len++]=d; } BigNum::BigNum(const char *s) //将一个字符串类型的变量转化为大数 { int t,k,index,L,i; memset(a,0,sizeof(a)); L=strlen(s); len=L/DLEN; if(L%DLEN) len++; index=0; for(i=L-1;i>=0;i-=DLEN) { t=0; k=i-DLEN+1; if(k<0) k=0; for(int j=k;j<=i;j++) t=t*10+s[j]-‘0‘; a[index++]=t; } } BigNum::BigNum(const BigNum &T):len(T.len) //拷贝构造函数 { int i; memset(a,0,sizeof(a)); for(i=0;i<len;i++) a[i]=T.a[i]; } BigNum & BigNum::operator=(const BigNum &n) //重载赋值运算符，大数之间赋值运算 { int i; len=n.len; memset(a,0,sizeof(a)); for(i=0;i<len;i++) a[i]=n.a[i]; return *this; } istream& operator>>(istream &in,BigNum &b) { char ch[MAXSIZE*4]; int i=-1; in>>ch; int L=strlen(ch); int count=0,sum=0; for(i=L-1;i>=0;) { sum=0; int t=1; for(int j=0;j<4&&i>=0;j++,i--,t*=10) { sum+=(ch[i]-‘0‘)*t; } b.a[count]=sum; count++; } b.len=count++; return in; } ostream& operator<<(ostream& out,BigNum& b) //重载输出运算符 { int i; cout<<b.a[b.len-1]; for(i=b.len-2;i>=0;i--) { printf("%04d",b.a[i]); } return out; } BigNum BigNum::operator+(const BigNum &T)const //两个大数之间的相加运算 { BigNum t(*this); int i,big; big=T.len>len?T.len:len; for(i=0;i<big;i++) { t.a[i]+=T.a[i]; if(t.a[i]>MAXN) { t.a[i+1]++; t.a[i]-=MAXN+1; } } if(t.a[big]!=0) t.len=big+1; else t.len=big; return t; } BigNum BigNum::operator-(const BigNum &T)const //两个大数之间的相减运算 { int i,j,big; bool flag; BigNum t1,t2; if(*this>T) { t1=*this; t2=T; flag=0; } else { t1=T; t2=*this; flag=1; } big=t1.len; for(i=0;i<big;i++) { if(t1.a[i]<t2.a[i]) { j=i+1; while(t1.a[j]==0) j++; t1.a[j--]--; while(j>i) t1.a[j--]+=MAXN; t1.a[i]+=MAXN+1-t2.a[i]; } else t1.a[i]-=t2.a[i]; } t1.len=big; while(t1.a[len-1]==0 && t1.len>1) { t1.len--; big--; } if(flag) t1.a[big-1]=0-t1.a[big-1]; return t1; } BigNum BigNum::operator*(const BigNum &T)const //两个大数之间的相乘 { BigNum ret; int i,j,up; int temp,temp1; for(i=0;i<len;i++) { up=0; for(j=0;j<T.len;j++) { temp=a[i]*T.a[j]+ret.a[i+j]+up; if(temp>MAXN) { temp1=temp-temp/(MAXN+1)*(MAXN+1); up=temp/(MAXN+1); ret.a[i+j]=temp1; } else { up=0; ret.a[i+j]=temp; } } if(up!=0) ret.a[i+j]=up; } ret.len=i+j; while(ret.a[ret.len-1]==0 && ret.len>1) ret.len--; return ret; } BigNum BigNum::operator/(const int &b)const //大数对一个整数进行相除运算 { BigNum ret; int i,down=0; for(i=len-1;i>=0;i--) { ret.a[i]=(a[i]+down*(MAXN+1))/b; down=a[i]+down*(MAXN+1)-ret.a[i]*b; } ret.len=len; while(ret.a[ret.len-1]==0 && ret.len>1) ret.len--; return ret; } int BigNum::operator%(const int &b)const //大数对一个 int类型的变量进行取模 { int i,d=0; for(i=len-1;i>=0;i--) d=((d*(MAXN+1))%b+a[i])%b; return d; } BigNum BigNum::operator^(const int &n)const //大数的n次方运算 { BigNum t,ret(1); int i; if(n<0)exit(-1); if(n==0)return 1; if(n==1)return *this; int m=n; while(m>1) { t=*this; for(i=1;(i<<1)<=m;i<<=1) t=t*t; m-=i; ret=ret*t; if(m==1) ret=ret*(*this); } return ret; } bool BigNum::operator>(const BigNum &T)const //大数和另一个大数的大小比较 { int ln; if(len>T.len) return true; else if(len==T.len) { ln=len-1; while(a[ln]==T.a[ln]&&ln>=0) ln--; if(ln>=0 && a[ln]>T.a[ln]) return true; else return false; } else return false; } bool BigNum::operator>(const int &t)const //大数和一个int类型的变量的大小比较 { BigNum b(t); return *this>b; } void BigNum::print() //输出大数 { int i; printf("%d",a[len-1]); for(i=len-2;i>=0;i--) printf("%04d",a[i]); printf("\n"); } BigNum a,d; BigNum op(int x) { BigNum a=x; BigNum s1=1; for(int i=1;i<=x;i++) s1=s1*i; return s1; } int n,m; int main() { int ca=1; //freopen("C:\\Users\\acer\\Desktop\\out.txt","w",stdout); while(scanf("%d%d",&m,&n)!=EOF&&(m||n)) //for(n=0;n<=100;n++) //for(m=0;m<=100;m++) { printf("Test #%d:\n",ca++); //(m+n)!*(m-n+1)/(m+1) if(m<n) { puts("0"); continue; } a=op(m+n); d=a*(m-n+1)/(m+1); d.print(); } }
/*
爆搜找规律
*/
#include<bits/stdc++.h>
using namespace std;
int n,m;
long long cur=0;
int ok(int x,int y)
{
if(x>n||y>m)
return 0;
return 1;
}
void dfs(int nown,int x,int y,int i)
// 现在售票处的50的张数，已经安排了的n和m的人数，现在安排第几位
{
//cout<<"x="<<x<<" y="<<y<<endl;
if(i==n+m)//将所有的人都安排好了，找到一组解
{
//cout<<"cur="<<cur<<endl;
cur++;
return ;
}
if(nown>0)//说明此时售票处有空余的50的
{
if(ok(x,y+1))//两种票都能用
dfs(nown-1,x,y+1,i+1);
if(ok(x+1,y))
dfs(nown+1,x+1,y,i+1);
}
else
{
//cout<<"x="<<x<<" y="<<y<<endl;
//cout<<"ok(x+1,y)="<<ok(x+1,y)<<endl;
if(ok(x+1,y))
{
dfs(nown+1,x+1,y,i+1);
//cout<<"ok"<<endl;
}
}
}
long long op(int x)
{
long long s=1;
for(int i=1;i<=x;i++)
s*=i;
return s;
}
int main()
{
freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin);
//freopen("C:\\Users\\acer\\Desktop\\out.txt","w",stdout);
for(n=0;n<=10;n++)
{
for(m=0;m<=n;m++)
{
cur=0;
dfs(0,0,0,0);
//cur=(cur*op(n)*op(m));
printf("%lld ",cur);
}
cout<<endl;
}
return 0;
}
/* 大数打表 */ #include <iostream> #include <stdio.h> #include <algorithm> #include <string.h> using namespace std; /* * 完全大数模板 * 输出cin>>a * 输出a.print(); * 注意这个输入不能自动去掉前导0的，可以先读入到char数组，去掉前导0，再用构造函数。 */ #define MAXN 9999 #define MAXSIZE 1010 #define DLEN 4 class BigNum { private: int a[500]; //可以控制大数的位数 int len; public: BigNum() { len=1;memset(a,0,sizeof(a)); } //构造函数 BigNum(const int); //将一个int类型的变量转化成大数 BigNum(const char*); //将一个字符串类型的变量转化为大数 BigNum(const BigNum &); //拷贝构造函数 BigNum &operator=(const BigNum &); //重载赋值运算符，大数之间进行赋值运算 friend istream& operator>>(istream&,BigNum&); //重载输入运算符 friend ostream& operator<<(ostream&,BigNum&); //重载输出运算符 BigNum operator+(const BigNum &)const; //重载加法运算符，两个大数之间的相加运算 BigNum operator-(const BigNum &)const; //重载减法运算符，两个大数之间的相减运算 BigNum operator*(const BigNum &)const; //重载乘法运算符，两个大数之间的相乘运算 BigNum operator/(const int &)const; //重载除法运算符，大数对一个整数进行相除 运算 BigNum operator^(const int &)const; //大数的n次方运算 int operator%(const int &)const; //大数对一个int类型的变量进行取模运算 bool operator>(const BigNum &T)const; //大数和另一个大数的大小比较 bool operator>(const int &t)const; //大数和一个int类型的变量的大小比较 void print(); //输出大数 }; BigNum::BigNum(const int b) //将一个int类型的变量转化为大数 { int c,d=b; len=0; memset(a,0,sizeof(a)); while(d>MAXN) { c=d-(d/(MAXN+1))*(MAXN+1); d=d/(MAXN+1); a[len++]=c; } a[len++]=d; } BigNum::BigNum(const char *s) //将一个字符串类型的变量转化为大数 { int t,k,index,L,i; memset(a,0,sizeof(a)); L=strlen(s); len=L/DLEN; if(L%DLEN) len++; index=0; for(i=L-1;i>=0;i-=DLEN) { t=0; k=i-DLEN+1; if(k<0) k=0; for(int j=k;j<=i;j++) t=t*10+s[j]-‘0‘; a[index++]=t; } } BigNum::BigNum(const BigNum &T):len(T.len) //拷贝构造函数 { int i; memset(a,0,sizeof(a)); for(i=0;i<len;i++) a[i]=T.a[i]; } BigNum & BigNum::operator=(const BigNum &n) //重载赋值运算符，大数之间赋值运算 { int i; len=n.len; memset(a,0,sizeof(a)); for(i=0;i<len;i++) a[i]=n.a[i]; return *this; } istream& operator>>(istream &in,BigNum &b) { char ch[MAXSIZE*4]; int i=-1; in>>ch; int L=strlen(ch); int count=0,sum=0; for(i=L-1;i>=0;) { sum=0; int t=1; for(int j=0;j<4&&i>=0;j++,i--,t*=10) { sum+=(ch[i]-‘0‘)*t; } b.a[count]=sum; count++; } b.len=count++; return in; } ostream& operator<<(ostream& out,BigNum& b) //重载输出运算符 { int i; cout<<b.a[b.len-1]; for(i=b.len-2;i>=0;i--) { printf("%04d",b.a[i]); } return out; } BigNum BigNum::operator+(const BigNum &T)const //两个大数之间的相加运算 { BigNum t(*this); int i,big; big=T.len>len?T.len:len; for(i=0;i<big;i++) { t.a[i]+=T.a[i]; if(t.a[i]>MAXN) { t.a[i+1]++; t.a[i]-=MAXN+1; } } if(t.a[big]!=0) t.len=big+1; else t.len=big; return t; } BigNum BigNum::operator-(const BigNum &T)const //两个大数之间的相减运算 { int i,j,big; bool flag; BigNum t1,t2; if(*this>T) { t1=*this; t2=T; flag=0; } else { t1=T; t2=*this; flag=1; } big=t1.len; for(i=0;i<big;i++) { if(t1.a[i]<t2.a[i]) { j=i+1; while(t1.a[j]==0) j++; t1.a[j--]--; while(j>i) t1.a[j--]+=MAXN; t1.a[i]+=MAXN+1-t2.a[i]; } else t1.a[i]-=t2.a[i]; } t1.len=big; while(t1.a[len-1]==0 && t1.len>1) { t1.len--; big--; } if(flag) t1.a[big-1]=0-t1.a[big-1]; return t1; } BigNum BigNum::operator*(const BigNum &T)const //两个大数之间的相乘 { BigNum ret; int i,j,up; int temp,temp1; for(i=0;i<len;i++) { up=0; for(j=0;j<T.len;j++) { temp=a[i]*T.a[j]+ret.a[i+j]+up; if(temp>MAXN) { temp1=temp-temp/(MAXN+1)*(MAXN+1); up=temp/(MAXN+1); ret.a[i+j]=temp1; } else { up=0; ret.a[i+j]=temp; } } if(up!=0) ret.a[i+j]=up; } ret.len=i+j; while(ret.a[ret.len-1]==0 && ret.len>1) ret.len--; return ret; } BigNum BigNum::operator/(const int &b)const //大数对一个整数进行相除运算 { BigNum ret; int i,down=0; for(i=len-1;i>=0;i--) { ret.a[i]=(a[i]+down*(MAXN+1))/b; down=a[i]+down*(MAXN+1)-ret.a[i]*b; } ret.len=len; while(ret.a[ret.len-1]==0 && ret.len>1) ret.len--; return ret; } int BigNum::operator%(const int &b)const //大数对一个 int类型的变量进行取模 { int i,d=0; for(i=len-1;i>=0;i--) d=((d*(MAXN+1))%b+a[i])%b; return d; } BigNum BigNum::operator^(const int &n)const //大数的n次方运算 { BigNum t,ret(1); int i; if(n<0)exit(-1); if(n==0)return 1; if(n==1)return *this; int m=n; while(m>1) { t=*this; for(i=1;(i<<1)<=m;i<<=1) t=t*t; m-=i; ret=ret*t; if(m==1) ret=ret*(*this); } return ret; } bool BigNum::operator>(const BigNum &T)const //大数和另一个大数的大小比较 { int ln; if(len>T.len) return true; else if(len==T.len) { ln=len-1; while(a[ln]==T.a[ln]&&ln>=0) ln--; if(ln>=0 && a[ln]>T.a[ln]) return true; else return false; } else return false; } bool BigNum::operator>(const int &t)const //大数和一个int类型的变量的大小比较 { BigNum b(t); return *this>b; } void BigNum::print() //输出大数 { int i; printf("%d",a[len-1]); for(i=len-2;i>=0;i--) printf("%04d",a[i]); printf("\n"); } BigNum a[101][101]; BigNum op(int x,int y) { BigNum a=x,b=y; BigNum s1=1,s2=1; for(int i=1;i<=x;i++) s1=s1*i; for(int i=1;i<=y;i++) s2=s2*i; return s1*s2; } int main() { for(int i=0;i<=100;i++) a[0][i]=0; for(int i=0;i<=100;i++) a[i][0]=1; op(100,100).print(); for(int i=1;i<=100;i++) { for(int j=1;j<=100;j++) { if(j<=i) { a[i][j]=a[i-1][j]+a[i][j-1]; a[i][j]=a[i][j]*op(i,j); } else a[i][j]=0; //a[i][j].print(); } } for(int i=0;i<=10;i++) { for(int j=0;j<=10;j++) { a[i][j].print(); cout<<" "; } cout<<endl; } return 0; }
标签：enter idt max lin auth pac not proc void
原文地址：http://www.cnblogs.com/wuwangchuxin0924/p/6142756.html
