HDU1517(大数运算)

发布时间：2020-12-14 04:10:25 所属栏目：大数据来源：网络整理

导读：A Multiplication Game Time Limit: 5000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 2540????Accepted Submission(s): 1452 Problem Description Stan and Ollie play the game of multiplication by multi

A Multiplication Game

Time Limit: 5000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2540????Accepted Submission(s): 1452

Problem Description

Stan and Ollie play the game of multiplication by multiplying an integer p by one of the numbers 2 to 9. Stan always starts with p = 1,does his multiplication,then Ollie multiplies the number,then Stan and so on. Before a game starts,they draw an integer 1 < n < 4294967295 and the winner is who first reaches p >= n.

Input

Each line of input contains one integer number n.

Output

For each line of input output one line either

Stan wins.

or

Ollie wins.

assuming that both of them play perfectly.

Sample Input

Sample Output

  
  
   
   Stan wins.
Ollie wins.
Stan wins.

#include<iostream>
#include<iomanip>
using namespace std;
int fab(int a[],int b[],int len,int c[])
{
?int i,temp=0;
?for(i=0;i<len;i++)
?{
??c[i]=(a[i]+b[i]+temp)%10000;
??temp=(a[i]+b[i]+temp)/10000;
?}
?if(temp!=0){c[i]=temp;i++;}
?return i;
}
void copy(int a[],int len)
{
?int i;
?for(i=0;i<len;i++)
?{
??a[i]=b[i];
?}
}

int main() { ?int a[110],b[110],c[110],n,m,len,j,i; ?cin>>n; ?while(n--) ?{ ??cin>>m; ??memset(a,sizeof(int)*110); ??memset(b,sizeof(int)*110); ??memset(c,sizeof(int)*110); ??a[0]=1;b[0]=1; ??len=1; ??if(m<3)cout<<"1"<<endl; ??else ??{ ???for(i=3;i<=m;i++) ???{ ????len=fab(a,b,c); ????copy(a,len); ????copy(b,c,len); ???} ???cout<<c[len-1]; ???for(i=len-2;i>=0;i--) ????cout<<setfill('0')<<setw(4)<<c[i]; ???cout<<endl; ??? ??} ?} ?return 0; }

（编辑：李大同）

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