题目:
http://acm.hdu.edu.cn/showproblem.php?pid=1005

代码:

 #include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int mod=;
struct matrix
{
int f[][];
}; matrix mul(matrix a,matrix b)
{
int i,j,k;
matrix c;
memset(c.f,,sizeof(c.f));
for(i=;i<;i++)
{
for(j=;j<;j++)
{
for(k=;k<;k++)
{
c.f[i][j]+=a.f[i][k]*b.f[k][j];
c.f[i][j]%=mod;
}
}
}
return c;
} matrix pow_mod(matrix a,int b)
{
matrix s;
int t;
memset(s.f,,sizeof(s.f));
for(int i=;i<;i++)
{
s.f[i][i]=;
}
while(b)
{
/*if(b&1)
{
s=mul(s,a);
}
a=mul(a,a);
b=b>>1;*/
t=b%;
b/=;
if(t!=)
s=mul(s,a);
a=mul(a,a);
}
return s;
}
int main()
{
int a,b,n;
while(scanf("%d%d%d",&a,&b,&n))
{
if(a==&&b==&&n==)
break;
matrix e;
e.f[][]=a;
e.f[][]=;
e.f[][]=b;
e.f[][]=;
e=pow_mod(e,n-);
printf("%d\n",(e.f[][]+e.f[][])%mod);
} return ;
}

hdu 1005 Number Sequence(矩阵连乘+二分快速求幂)的更多相关文章

  1. HDU - 1005 Number Sequence 矩阵快速幂

    HDU - 1005 Number Sequence Problem Description A number sequence is defined as follows:f(1) = 1, f(2 ...

  2. HDU 1005 Number Sequence(矩阵快速幂,快速幂模板)

    Problem Description A number sequence is defined as follows: f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1 ...

  3. HDU - 1005 -Number Sequence(矩阵快速幂系数变式)

    A number sequence is defined as follows:  f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) m ...

  4. HDU 1005 Number Sequence(数列)

    HDU 1005 Number Sequence(数列) Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Jav ...

  5. HDU 1005 Number Sequence(数论)

    HDU 1005 Number Sequence(数论) Problem Description: A number sequence is defined as follows:f(1) = 1, ...

  6. HDU 1005 Number Sequence【斐波那契数列/循环节找规律/矩阵快速幂/求(A * f(n - 1) + B * f(n - 2)) mod 7】

    Number Sequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)T ...

  7. HDU 1005 Number Sequence(矩阵)

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission( ...

  8. HDU - 1005 Number Sequence (矩阵快速幂)

    A number sequence is defined as follows: f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mo ...

  9. HDU 1005 Number Sequence【多解,暴力打表,鸽巢原理】

    Number Sequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)T ...

随机推荐

  1. STM8S103-STVD建立汇编代码项目

    转载:http://blog.csdn.net/u010093140/article/details/49983397 STVD本来就比较少人用,STVD汇编就更少人用了,不过STM8汇编我自己还是满 ...

  2. jq不懂的地方

    在循环列表中,获取input标签的值,不能用id获取,用class获取值,通过父级属性找到class,this 指当前点击的位置var UID = $(this).parents("tr&q ...

  3. Linux重新命名文件夹

    linux 重命名文件和文件夹   linux下重命名文件或文件夹的命令mv既可以重命名,又可以移动文件或文件夹. 例子:将目录A重命名为B mv A B 例子:将/a目录移动到/b下,并重命名为c ...

  4. POJ 3093 Margaritas on the River Walk(背包)

    题意 n个有体积的物品,问选取一些物品,且不能再继续选有多少方法? n<=1000 题解 以前的考试题.当时是A了,但发现是数据水,POJ上WA了. 把体积从小到大排序枚举没选的物品中体积最小的 ...

  5. [POJ2823][洛谷P1886]滑动窗口 Sliding Window

    题目大意:有一列数,和一个窗口,一次能框连续的s个数,初始时窗口在左端,不断往右移动,移到最右端为止,求每次被框住的s个数中的最小数和最大数. 解题思路:这道题是一道区间查询问题,可以用线段树做.每个 ...

  6. js各种验证

    1. var Validate = function() { //账号验证 字母,数字,下划线,不能少于6位大于20位 this.isName =function(value){ var reg = ...

  7. curl命令查看响应时间

    curl -w "%{time_namelookup}::%{time_connect}::%{time_starttransfer}::%{time_total}::%{speed_dow ...

  8. 【codeforces 816C】Karen and Game

    [题目链接]:http://codeforces.com/contest/816/problem/C [题意] 给你一个n*m的矩阵; 一开始所有数字都是0; 每次操作,你能把某一行,或某一列的数字全 ...

  9. POJ——T 1422 Air Raid

    http://poj.org/problem?id=1422 Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 8579   A ...

  10. uva live 2326 - Moving Tables

    把房间号映射在一条坐标上,然后排序,最后找从左到右找一次可行的计划,最后找从左到右找一次可行的计划,最后找从左到右找一次可行的计划,最后找从左到右找一次可行的计划, ............ 次数*1 ...