hdu4704 Sum 2013 Multi-University Training Contest 10 数论题

Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 127    Accepted Submission(s): 60

Problem Description

 

Sample Input

2

 

Sample Output

2

解题思路：

很容易看得出，这是个组合数学的插板问题，答案为2^(n-1);

由于n特别大，则(2^(n-1))%1000000007=(2^((n-1)%1000000006))%1000000007;

因为1000000007为素数，设为p，有2^p=2(%p)，即2^(p-1)=1(%p);

用个大数对小数取模，和快速幂就好了

 #include <iostream>
 #include <stdio.h>
 #include <algorithm>
 #include <stdlib.h>
 #include <string.h>
 #include <queue>
 #define ll long long int
 #define mod 1000000007
 #define INF 1000000006
 using namespace std;
 char a[];
 int b[];
 int t;
 ll fun()
 {
     int i,j;
     ll sum=;
     for(i=t;i>=;i--)
     {
         sum=sum*+b[i];
         sum%=INF;
     }
     return sum;
 }
 ll funa(ll sum)
 {
     if(sum==)
     return ;
     if(sum==)
     return ;
     if(sum&)
     {
         ll temp=funa((sum-)/);
         return temp*temp%mod*%mod;
     }
     else
     {
         ll temp=funa(sum/);
         return temp*temp%mod;
     }
 }
 int main()
 {
     while(scanf("%s",a)!=EOF)
     {
         t=strlen(a);
         int i,j;
         j=;
         memset(b,,sizeof(b));
         for(i=t-;i>=;i--)
         {
             b[j++]=a[i]-'';
         }
         for(i=;i<j;i++)
         if(b[i]>) {b[i]--;break;}
         else b[i]=;
         ll res=fun();
         cout<<funa(res)<<endl;
     }
 }