HDU 5451 Best Solver 数论 快速幂 2015沈阳icpc

Best Solver

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 65535/102400 K (Java/Others)
Total Submission(s): 1115    Accepted Submission(s): 645


Problem Description
The so-called best problem solver can easily solve this problem, with his/her childhood sweetheart.

It is known that  y=(5+26√)1+2x.
For a given integer x (0x<232) and a given prime number M (M46337), print [y]%M. ([y] means the integer part of y)
 
Input
An integer  T (1<T1000), indicating there are T test cases.
Following are T lines, each containing two integers x and M, as introduced above.
 
Output
The output contains exactly  T lines.
Each line contains an integer representing [y]%M.
 
Sample Input
7 0 46337 1 46337 3 46337 1 46337 21 46337 321 46337 4321 46337
 
Sample Output
Case #1: 97 Case #2: 969 Case #3: 16537 Case #4: 969 Case #5: 40453 Case #6: 10211 Case #7: 17947
 
Source
 
Recommend
wange2014
 

题目描述:

  对于,给出x和mod,求y向下取整后取余mod的值为多少?

分析:

  这种(a+sqrt(b))^n%mod,向下取整取余的式子有个递推式   S(n) = 2*a*S(n-1)+(b-a*a)*S(n-2) 这个式子得到的结果是向上取整,我们这里是向下取整,所以最后得到的结果还要减一

  我这篇博客https://www.cnblogs.com/l609929321/p/9398349.html有这个递推式的推导过程

  由上诉递推式不难得到:S(n) = (10*S(n-1)-S(n-2)+mod)%mod

  但是这个题目的指数1+2^x(x<2^32)非常大,就算用矩阵快速幂也不行,所以我们得对式子进行一次优化

  看到指数很大的类似斐波拉数列问题(都是线性递推式)很容易的想到找循环节

  因为mod<=46337,所以我们直接遍历求前面46337项一定会出现循环节,然后记录下循环节的位置就好

  再对指数按照循环节的位置进行取模就可以了

AC代码

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#define ls (r<<1)
#define rs (r<<1|1)
#define debug(a) cout << #a << " " << a << endl
using namespace std;
typedef long long ll;
const ll maxn = 1e5 + 10;
const ll mod = 1e9 + 7;
ll r[maxn], ans[maxn], x, m;
ll qow( ll b, ll n, ll mod ) {
    ll ans = 1;
    while( n ) {
        if( n&1 ) {
            ans = ans*b%mod;
        }
        b = b*b%mod;
        n /= 2;
    }
    return ans;
}
int main() {
    ll T, t = 0;
    cin >> T;
    while( T -- ) {
        ll x, m;
        t ++;
        cin >> x >> m;
        ans[0] = 2, ans[1] = 10;
        for( ll i = 2; i < maxn; i ++ ) {
            ans[i] = (10*ans[i-1]-ans[i-2]+m)%m;  //根据类似斐波拉数得到的一个递推式
            if( ans[i-1] == ans[0] && ans[i] == ans[1] ) { //因为x很大,这里计算了循环周期
                r[m] = i-1;
                break;
            }
        }
        ll k = (1+qow(2,x,r[m]))%r[m];
        cout << "Case #" << t << ": " << (ans[k]-1+m)%m << endl;
    }
	return 0 ;
}

  

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转载自www.cnblogs.com/l609929321/p/9398442.html