Buy the Ticket
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4726 Accepted Submission(s): 1993
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.
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.
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.
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.
Sample Input
3 0 3 1 3 3 0 0
Sample Output
Test #1: 6 Test #2: 18 Test #3: 180
Author
HUANG, Ninghai
Recommend
当然这是一道卡特兰数的变式题
递推公式很容易得出
F[i][j]=F[i-1][j]+F[i][j-1];(当i<j时 F[i][j]=0)
最终结果是 F[m][n]*m!*n! 显然涉及高精度
高精度是这题的主角
一：不压位+F[m][n]+N!+高精乘高精
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <ctime>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <string>
#define oo 0x13131313
using namespace std;
int C[102][102][50];
int len[102][102];
int JC[102][200];
int LEN[102];
void getadd(int a,int b)
{
int a1=a,a2=a-1,b1=b-1,b2=b;
int temp=0;
len[a][b]=max(len[a1][b1],len[a2][b2])+1;
for(int i=0;i<len[a][b];i++)
{
temp+=C[a1][b1][i]+C[a2][b2][i];
C[a][b][i]=temp%10;
temp=temp/10;
}
if(C[a][b][len[a][b]-1]==0) len[a][b]--;
}
void getC()
{
C[1][1][0]=1;
len[1][1]=1;
for(int i=1;i<=100;i++)
{
C[i][0][0]=1;
len[i][0]=1;
}
for(int j=1;j<=100;j++)
for(int i=1;i<=100;i++)
{
len[i][j]=1;
if(j<=i)
getadd(i,j);
}
}
void getx(int X)
{
double k=(log((double)X)/log(10.0));
int K=(int)(k+1);
LEN[X]=LEN[X-1]+K;
int temp=0;
for(int i=0;i<LEN[X];i++)
{
temp+=JC[X-1][i]*X;
JC[X][i]=temp%10;
temp=temp/10;
}
if(JC[X][LEN[X]-1]==0) LEN[X]--;
}
void getJC()
{
JC[0][0]=1;LEN[0]=1;JC[1][0]=1;LEN[1]=1;
for(int i=1;i<=100;i++)
getx(i);
}
void highXhigh(int *c,int &lenc,int *a,int lena,int *b,int lenb)
{
lenc=lena+lenb;
int temp=0;
for(int i=0;i<lena;i++)
for(int j=0,temp=0;j<=lenb;j++)
{
temp+=a[i]*b[j]+c[i+j];
c[i+j]=temp%10;
temp=temp/10;
}
if(c[lenc-1]==0) lenc--;
}
int buffer1[500],buffer2[500],len1,len2;
void getans(int n,int m)
{
memset(buffer1,0,sizeof(buffer1));
memset(buffer2,0,sizeof(buffer2));
highXhigh(buffer1,len1,JC[n],LEN[n],C[n][m],len[n][m]);
highXhigh(buffer2,len2,buffer1,len1,JC[m],LEN[m]);
for(int i=len2-1;i>=0;i--)
{
printf("%d",buffer2[i]);
}
}
int main()
{
// freopen("a.in","r",stdin);
// freopen("a.out","w",stdout);
getC();
getJC();
int Case=0,n,m;
while(scanf("%d%d",&n,&m)!=EOF &&(n!=0||m!=0))
{
Case++;
printf("Test #%d:
",Case);
if(n>=m)
getans(n,m);
else printf("0");
printf("
");
}
return 0;
}
有几点要注意的
#71 j<=lenb 以及所有的估计长度都比实际
可能多一位的原因是保证 temp 最后为0；
二：压位+F[m][n]+N!+高精乘高精
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <ctime>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <string>
#define oo 0x13131313
#define B 10000
using namespace std;
int C[102][102][30];
int len[102][102];
int JC[102][50];
int LEN[102];
void getadd(int a,int b)
{
int a1=a,a2=a-1,b1=b-1,b2=b;
int temp=0;
len[a][b]=max(len[a1][b1],len[a2][b2])+1;
for(int i=0;i<len[a][b];i++)
{
temp+=C[a1][b1][i]+C[a2][b2][i];
C[a][b][i]=temp%B;
temp=temp/B;
}
if(C[a][b][len[a][b]-1]==0) len[a][b]--;
}
void getC()
{
C[1][1][0]=1;
len[1][1]=1;
for(int i=1;i<=100;i++)
{
C[i][0][0]=1;
len[i][0]=1;
}
for(int j=1;j<=100;j++)
for(int i=1;i<=100;i++)
{
len[i][j]=1;
if(j<=i)
getadd(i,j);
}
}
void getx(int X)
{
// double k=(log((double)X)/log(10.0));
// int K=(int)(k+1);
LEN[X]=LEN[X-1]+1;
int temp=0;
for(int i=0;i<LEN[X];i++)
{
temp+=JC[X-1][i]*X;
JC[X][i]=temp%B;
temp=temp/B;
}
if(JC[X][LEN[X]-1]==0) LEN[X]--;
}
void getJC()
{
JC[0][0]=1;LEN[0]=1;JC[1][0]=1;LEN[1]=1;
for(int i=1;i<=100;i++)
getx(i);
}
void highXhigh(int *c,int &lenc,int *a,int lena,int *b,int lenb)
{
lenc=lena+lenb;
int temp=0;
for(int i=0;i<lena;i++)
for(int j=0,temp=0;j<=lenb;j++)
{
temp+=a[i]*b[j]+c[i+j];
c[i+j]=temp%B;
temp=temp/B;
}
if(c[lenc-1]==0) lenc--;
}
int buffer1[120],buffer2[120],len1,len2;
void getans(int n,int m)
{
memset(buffer1,0,sizeof(buffer1));
memset(buffer2,0,sizeof(buffer2));
highXhigh(buffer1,len1,JC[n],LEN[n],C[n][m],len[n][m]);
highXhigh(buffer2,len2,buffer1,len1,JC[m],LEN[m]);
if(len2-1>=0) printf("%d",buffer2[len2-1]);
for(int i=len2-2;i>=0;i--)
{
printf("%04d",buffer2[i]);
}
}
int main()
{
freopen("a.in","r",stdin);
freopen("b.out","w",stdout);
getC();
getJC();
int Case=0,n,m;
while(scanf("%d%d",&n,&m)!=EOF &&(n!=0||m!=0))
{
Case++;
printf("Test #%d:
",Case);
if(n>=m)
getans(n,m);
else printf("0");
printf("
");
}
return 0;
}
改压位十分简单
把预处理的define B 改一下就好了
随便压几位都随便改
不过要注意的是
乘以非高精数的时候注意计算他的长度/4