【矩阵快速幂】HDU_4565_So Easy!-优快云博客

本文介绍了一种使用矩阵快速幂解决特定数列问题的方法。针对给定的数列公式，通过矩阵运算加速大数指数计算，并提供了完整的C++代码实现。适用于解决形如Sn=a^n+b^n%m的问题。

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So Easy!

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5051 Accepted Submission(s): 1660

Problem Description

　　A sequence S _n 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 S _n.
　　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 < 2 ¹⁵, (a-1) ²< b < a ², 0 < b, n < 2 ³¹.The input will finish with the end of file.

Output

　　For each the case, output an integer S _n.

Sample Input

Sample Output

Source

2013 ACM-ICPC长沙赛区全国邀请赛——题目重现

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zhoujiaqi2010

/*化一化就转化成a^n+b^n%m的问题
跟lightOJ1070一样
*/
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn=2;
typedef struct node{
    LL a[maxn][maxn];
    node(){
        memset(a,0,sizeof(a));
    }
}Matrix;
Matrix multiply(Matrix x,Matrix y,LL mod){
    Matrix ans;
    for(int i=0;i<maxn;i++){
        for(int j=0;j<maxn;j++){
            for(int k=0;k<maxn;k++){
                ans.a[i][j]=(ans.a[i][j]+x.a[i][k]*y.a[k][j])%mod;
            }
        }
    }
    return ans;
}
Matrix quickpow(Matrix x,int a,int mod){
    Matrix ans;
    for(int i=0;i<maxn;i++)
        ans.a[i][i]=1;
    while(a){
        if(a&1)
            ans=multiply(ans,x,mod);
        x=multiply(x,x,mod);
        a>>=1;
    }
    return ans;
}
int main(){
    LL a,b,m;
    int n;
    while(~scanf("%I64d%I64d%d%I64d",&a,&b,&n,&m)){
        if(n<3){
            if(n==0) printf("%I64d\n",2%m);
            else if(n==1) printf("%I64d\n",(2*a)%m);
            else if(n==2) printf("%I64d\n",(2*a*a+2*b)%m);
            continue;
        }
        Matrix base,x;
        base.a[0][0]=2*a*a+2*b;base.a[0][1]=2*a;
        x.a[0][0]=2*a%m;x.a[0][1]=1;x.a[1][0]=(b-a*a%m+m)%m;x.a[1][1]=0;//没取模就WA了
        Matrix ans=multiply(base,quickpow(x,n-2,m),m);
        printf("%I64d\n",ans.a[0][0]);
    }
}