hdu 4565 So Easy!

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 20132 3 2 20132 2 1 2013

 

Sample Output

4144題目分析： 
題目要求這個得值
但是模數前有更號，所以無法直接計算，我們發現
0< a, m < 215, (a-1)2< b < a2, 0 < b, n < 231
所以 0 <| a+sqrt( b ) | < 1
可得運算式：，由二項式展開可知等號右邊一坨是整數並且加的數小於一，所以等式成立
然後我們設 Kn 為為等號的左邊，將運算式化為遞推形式後，再利用矩陣連乘來解決 Kn 的問題
轉化過程就是移兩次項，每次都將指數約去即可化簡
# include<iostream># include<cstdio># include<cstring>typedef __int64 ll;using namespace std;ll n,m,a,b;struct node{   ll x[2][2];} op;node cheng(node a,node b){    node t;    int i,j,k;    for (i=0;i<2;i++)    for (j=0;j<2;j++)    {        ll sum=0;        for (k=0;k<2;k++)        sum=(sum+(ll)a.x[i][k]*b.x[k][j])%m;        t.x[i][j]=sum;    }    return t;}int main(){    while (~scanf("%I64d%I64d%I64d%I64d",&a,&b,&n,&m))    {        ll x,y;        x=(2*a)%m; y=(2*(a*a+b))%m;        op.x[0][0]=(2*a)%m; op.x[0][1]=1;        op.x[1][0]=((b-a*a%m)+m)%m;//不要掉了這裡的+m         op.x[1][1]=0;        if (n==1) { printf("%I64d\n",x); continue;};        if (n==2) { printf("%I64d\n",y); continue;};        node ans;        ans=op;        n=n-3;        while (n)        {            if (n&1) op=cheng(op,ans);            ans=cheng(ans,ans);            n=n/2;        }        printf("%I64d\n",(op.x[0][0]*y+op.x[1][0]*x)%m);    }    return 0;}