Train Problem II |
| Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) |
| Total Submission(s): 138 Accepted Submission(s): 80 |
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Problem Description
As we all know the Train Problem I, the boss of the Ignatius Train Station want to know if all the trains come in strict-increasing order, how many orders that all the trains can get out of the railway.
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Input
The input contains several test cases. Each test cases consists of a number N(1<=N<=100). The input is terminated by the end of file.
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Output
For each test case, you should output how many ways that all the trains can get out of the railway. |
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Sample Input
1 2 3 10 |
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Sample Output
1 2 5 16796
Hint
The result will be very large, so you may not process it by 32-bit integers.
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#include"stdio.h"
#include"string.h"
int ch[101][200]={0};
int temp[200]={0};
int add(int* x,int* y)
{
int i;
for(i=0;x[i]!=0&&y[i]!=0;i++)
{
x[i]+=y[i]-'0';
}
for(i;y[i]!=0;i++)
x[i]+=y[i];
int t;
for(t=0;x[t]!=0;t++)
{
if(x[t+1]!=0)
x[t+1]+=(x[t]-'0')/10;
else
{
if((x[t]-'0')/10!=0)
x[t+1]+=(x[t]-'0')/10+'0';
else
x[t+1]=0;
}
x[t]=(x[t]-'0')%10+'0';
}
return t;
}
int* mul(int* x,int* y)
{
for(int i=0;i<200;i++)
temp[i]=0;
for(int i=0;x[i]!=0;i++)
for(int j=0;y[j]!=0;j++)
{
if(temp[i+j]!=0)
temp[i+j]+=(x[i]-'0')*(y[j]-'0');
else
temp[i+j]+=(x[i]-'0')*(y[j]-'0')+'0';
}
int length=0;
while(temp[length]!=0)
length++;
for(int t=0;t<length;t++)
{
temp[t+1]+=(temp[t]-'0')/10;
temp[t]=(temp[t]-'0')%10+'0';
}
if(temp[length]!=0)
temp[length]+='0';
while(temp[length]!=0)
{
if((temp[length]-'0')/10!=0)
temp[length+1]+=temp[length]/10+'0';
temp[length]=(temp[length]-'0')%10+'0';
length++;
}
return temp;
}
int main()
{
ch[0][0]='1';
ch[1][0]='1';
for(int i=2;i<=100;i++)
{
for(int j=0;j<i;j++)
{
add(ch[i],mul(ch[j],ch[i-j-1]));
}
}
int n,t=199;
while(scanf("%d",&n)>0)
{
t=199;
while(t>=0)
{
if(ch[n][t]!=0)
printf("%c",ch[n][t]);
t--;
}
printf("\n");
}
return 0;
}
本文转自NewPanderKing51CTO博客，原文链接：http://www.cnblogs.com/newpanderking/archive/2011/07/29/2121146.html ，如需转载请自行联系原作者
