Problem of Precision

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1957 Accepted Submission(s): 1206

Problem Description

Input

The first line of input gives the number of cases, T. T test cases follow, each on a separate line. Each test case contains one positive integer n. (1 <= n <= 10^9)

Output

For each input case, you should output the answer in one line.

Sample Input

3 1 2 5

Sample Output

9 97 84

Source

HDOJ 2008 Summer Exercise（4）- Buffet Dinner

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lcy

So Easy!

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6238 Accepted Submission(s): 2064

Problem Description

　　A sequence Sn is defined as:

Where a, b, n, m are positive integers.┌x┐is the ceil of x. For example, ┌3.14┐=4. You are to calculate Sn.
　　You, a top coder, say: So easy!

Input

　　There are several test cases, each test case in one line contains four positive integers: a, b, n, m. Where 0< a, m < 215, (a-1)2< b < a2, 0 < b, n < 231.The input will finish with the end of file.

Output

　　For each the case, output an integer Sn.

Sample Input

2 3 1 2013 2 3 2 2013 2 2 1 2013

Sample Output

4 14 4

Source

2013 ACM-ICPC长沙赛区全国邀请赛——题目重现

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zhoujiaqi2010

题目大意：两道题目几乎一样，都是给以个带根号的式子，让你求这个式子的n次幂对一个数取模。

解题思路：

这样的式子与斐波那契的通项公式有一些相似。

对于第一题。（5+2sqrt（6））^n

考虑该式子能否化简成fn = an+bn*sqrt(6)

通项公式为 fn = an + bn*sqrt(6);

则fn-1 = an-1 + bn-1*sqrt（6）；

fn=5+sqrt(6)*fn-1

得到

an = 5*an-1+12*bn-1

bn=5*an-1+2*bn-1

所以通过矩阵快速幂可以求得an，

然后再利用5-2*sqrt（6）的通项公式消掉bn即可

HDU2256

#include<bits/stdc++.h>
#define rep(i,j,k) for(int i=j;i<=k;i++)
#define sca(x) scanf("%d",&x);
#define LL long long
using namespace std;
const int mod = 1024;
struct node
{
    int a[3][3];
};

void Print(node k)
{
    for(int i=1;i<=2;i++)
    {
        for(int j=1;j<=2;j++)
        {
            cout<<k.a[i][j]<<" ";
        }
        cout<<endl;
    }
}

node mul(node x,node y)
{
    node tmp;
    memset(tmp.a,0,sizeof(tmp.a));
    for(int i=1;i<=2;i++)
    {
        for(int j=1;j<=2;j++)
        {
            for(int k=1;k<=2;k++)
            {
                tmp.a[i][j]+=(x.a[i][k]*y.a[k][j])%mod;
            }
            tmp.a[i][j]%=mod;
        }
    }
    //Print(tmp);
    return tmp;
}


void solve(int k)
{
    node ans,tmp;
    memset(ans.a,0,sizeof(ans.a));
    memset(tmp.a,0,sizeof(tmp.a));
    ans.a[1][1]=5;ans.a[2][1]=2;
    tmp.a[1][1]=5,tmp.a[1][2]=12;
    tmp.a[2][1]=2,tmp.a[2][2]=5;
    while(k)
    {
        if(k&1)
            ans=mul(tmp,ans);
        tmp=mul(tmp,tmp);
        k>>=1;
    }
    printf("%d\n",(2*ans.a[1][1]-1)%mod);
}
int main()
{
    int t;
    sca(t);
    while(t--)
    {
        int n;
        sca(n);
        solve(n-1);
    }
}

#include<bits/stdc++.h>
#define rep(i,j,k) for(int i=j;i<=k;i++)
#define sca(x) scanf("%d",&x);
#define LL long long
using namespace std;

LL mod = 1024;
struct node
{
    LL a[3][3];
};


node mul(node x,node y)
{
    node tmp;
    memset(tmp.a,0,sizeof(tmp.a));
    for(int i=1;i<=2;i++)
    {
        for(int j=1;j<=2;j++)
        {
            for(int k=1;k<=2;k++)
            {
                tmp.a[i][j]+=(x.a[i][k]*y.a[k][j])%mod;
            }
            tmp.a[i][j]%=mod;
        }
    }
    //Print(tmp);
    return tmp;
}

LL aa,bb;
void solve(int k)
{
    node ans,tmp;
    memset(ans.a,0,sizeof(ans.a));
    memset(tmp.a,0,sizeof(tmp.a));
    ans.a[1][1]=aa;ans.a[2][1]=1;
    tmp.a[1][1]=aa,tmp.a[1][2]=bb;
    tmp.a[2][1]=1,tmp.a[2][2]=aa;
    while(k)
    {
        if(k&1)
            ans=mul(tmp,ans);
        tmp=mul(tmp,tmp);
        k>>=1;
    }
    printf("%d\n",(2*ans.a[1][1])%mod);
}
int main()
{
    int n;
    while(~scanf("%lld%lld%d%lld",&aa,&bb,&n,&mod))
    {
        solve(n-1);
    }
}

hdu 2256 4565 （矩阵快速幂递推）

Problem of Precision

So Easy!

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hdu 2256 4565 （矩阵快速幂递推）