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[100010];
int b[100010];
int t;
ll fun()
{
    int i,j;
    ll sum=0;
    for(i=t;i>=0;i--)
    {
        sum=sum*10+b[i];
        sum%=INF;
    }
    return sum;
}
ll funa(ll sum)
{
    if(sum==0)
    return 1;
    if(sum==1)
    return 2;
    if(sum&1)
    {
        ll temp=funa((sum-1)/2);
        return temp*temp%mod*2%mod;
    }
    else
    {
        ll temp=funa(sum/2);
        return temp*temp%mod;
    }
}
int main()
{
    while(scanf("%s",a)!=EOF)
    {
        t=strlen(a);
        int i,j;
        j=0;
        memset(b,0,sizeof(b));
        for(i=t-1;i>=0;i--)
        {
            b[j++]=a[i]-'0';
        }
        for(i=0;i<j;i++)
        if(b[i]>0) {b[i]--;break;}
        else b[i]=9;
        ll res=fun();
        cout<<funa(res)<<endl;
    }
}