HDU1517(大数运算)

最新推荐文章于 2020-05-04 11:11:13 发布

晓风残月xj

最新推荐文章于 2020-05-04 11:11:13 发布

阅读量796

点赞数

分类专栏：数学-大数处理文章标签：大数处理

本文链接：https://blog.youkuaiyun.com/xiaofengcanyuexj/article/details/8883818

版权

数学-大数处理专栏收录该内容

13 篇文章

订阅专栏

本文深入探讨了游戏开发中的核心策略与算法实现，包括数值分析、数据结构、算法优化等关键技术，旨在提升游戏性能与用户体验。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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 b[],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,0,sizeof(int)*110);
  memset(b,0,sizeof(int)*110);
  memset(c,0,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,len,c);
    copy(a,b,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;
}