HDU4686 Arc of Dream 矩阵快速幂
Arc of Dream
Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 4246 Accepted Submission(s): 1332

where
a0 = A0
ai = ai-1*AX+AY
b0 = B0
bi = bi-1*BX+BY
What is the value of AoD(N) modulo 1,000,000,007?
Each test case contains 7 nonnegative integers as follows:
N
A0 AX AY
B0 BX BY
N is no more than 1018, and all the other integers are no more than 2×109.
bi = bi-1*BX+BY
a[i]*b[i] = (a[i-1]*Ax+Ay)(b[i-1]*Bx+By)
= Ax*Bx*a[i-1]*b[i-1]+Ay*Bx*b[i-1]+Ax*By*a[i-1]+Ay*By
s
[s[n-1],a[n-2]*b[n-2], b[n-2], a[n-2], 1]
A
[1 ,0 ,0 ,0 ,0]
[Ax*Bx ,Ax*Bx ,0 ,0 ,0]
[Ay*Bx ,Ay*Bx ,Bx ,0 ,0]
[Ax*By ,Ax*By ,0 ,Ax ,0]
[Ay*By ,Ay*By ,By ,Ay ,1]//n-1
s
[s[n], a[n-1]*b[n-1], b[n-1], a[n-1], 1]
#include<bits/stdc++.h>
#define N 5
#define mes(x) memset(x, 0, sizeof(x));
#define ll long long
const ll mod = 1e9+;
const int MAX = 0x7ffffff;
using namespace std;
struct mat {
ll a[N][N];
mat() {
memset(a, , sizeof(a));
}
mat operator * (mat b) {
mat c;
for (int i = ; i < N; i++)
for (int j = ; j < N; j++)
for (int k = ; k < N; k++)
c.a[i][j] = (c.a[i][j] + a[i][k] * b.a[k][j]) % mod;
return c;
}
};
mat f(mat b, ll m) {
mat c;
for (int i = ; i < N; i++)
c.a[i][i] = ;
while (m) {
if (m & )
c = c * b;
b = b * b;
m >>= ;
}
return c;
}
int main()
{
ll n, A0,Ax, Ay, Bx,By,B0;
mat A, s;
while(~scanf("%lld", &n)){
scanf("%lld%lld%lld%lld%lld%lld", &A0, &Ax, &Ay, &B0, &Bx, &By);
if(n == ){
printf("0\n");
continue;
}
mes(A.a);
mes(s.a);
s.a[][] = s.a[][] = (A0%mod*B0%mod)%mod;
s.a[][] = B0%mod;
s.a[][] = A0%mod;
s.a[][] = ;
A.a[][] = ;
A.a[][] = A.a[][] = (Ax%mod*Bx%mod)%mod;A.a[][] = Bx%mod;
A.a[][] = A.a[][] = (Ay%mod*Bx%mod)%mod;A.a[][] = Ax%mod;
A.a[][] = A.a[][] = (Ax%mod*By%mod)%mod;
A.a[][] = A.a[][] = (Ay%mod*By%mod)%mod;
A.a[][] = By;
A.a[][] = Ay;
A.a[][] = ;
A = f(A,n-);
s = s*A;
printf("%lld\n", (mod+s.a[][])%mod);
}
}
HDU4686 Arc of Dream 矩阵快速幂的更多相关文章
- HDU4686——Arc of Dream矩阵快速幂
题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=4686 题目大意: 已知a0=A0, ai=Ax*ai-1+Ay; b0=B0, bi=Bx*bi-1 ...
- HDU4686 Arc of Dream —— 矩阵快速幂
题目链接:https://vjudge.net/problem/HDU-4686 Arc of Dream Time Limit: 2000/2000 MS (Java/Others) Memo ...
- hdu----(4686)Arc of Dream(矩阵快速幂)
Arc of Dream Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)Tota ...
- S - Arc of Dream 矩阵快速幂
An Arc of Dream is a curve defined by following function: where a 0 = A0 a i = a i-1*AX+AY b 0 = B0 ...
- hdu 4686 Arc of Dream(矩阵快速幂)
链接:http://acm.hdu.edu.cn/showproblem.php?pid=4686 题意: 其中a0 = A0ai = ai-1*AX+AYb0 = B0bi = bi-1*BX+BY ...
- HDU 4686 Arc of Dream 矩阵快速幂,线性同余 难度:1
http://acm.hdu.edu.cn/showproblem.php?pid=4686 当看到n为小于64位整数的数字时,就应该有个感觉,acm范畴内这应该是道矩阵快速幂 Ai,Bi的递推式题目 ...
- HDOJ 4686 Arc of Dream 矩阵高速幂
矩阵高速幂: 依据关系够建矩阵 , 高速幂解决. Arc of Dream Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 65535/ ...
- HDU4686 Arc of Dream 矩阵
欢迎访问~原文出处——博客园-zhouzhendong 去博客园看该题解 题目传送门 - HDU4686 题意概括 a0 = A0 ai = ai-1*AX+AY b0 = B0 bi = bi-1* ...
- hdu 4686 Arc of Dream_矩阵快速幂
题意:略 构造出矩阵就行了 | AX 0 AXBY AXBY 0 | ...
随机推荐
- PHP生成二维码【谷歌API+qrcode+圆角Logo】
方法一:谷歌二维码API 接口地址:https://chart.googleapis.com/chart 官方文档:https://developers.google.com/chart/infogr ...
- Fragement
package com.exmple.frage; import java.util.ArrayList; import java.util.Calendar; import java.util.Ha ...
- Java随学随记
1.一个Java源文件可包含三个“顶级”要素: (1)一个包(package)声明(可选) (2)任意数量的导入(import)语句 (3)类(class)声明 该三要素必须以上顺序出现.即,任何导入 ...
- Java简单数据类型转换
1. Integer<---String (1) Integer x = new Integer(Integer.parseInt(String)); 2. Integer<--- ...
- Sublime Text设置快捷键让html文件在浏览器打开
一.安装View In Browser插件 快捷键 Ctrl+Shift+P(菜单栏Tools->Command Paletter),输入 pcip选中Install Package并回车,输入 ...
- du -sh
评估文件空间利用率: [root@vm-xiluhua][/home]$ du -sh /home 409M /home --exclude选项,排除指定模式的文件的大小 [root@vm-xiluh ...
- JavaEE基础(二十)/IO流
1.IO流(IO流概述及其分类) 1.概念 IO流用来处理设备之间的数据传输 Java对数据的操作是通过流的方式 Java用于操作流的类都在IO包中 流按流向分为两种:输入流,输出流. 流按操作类型分 ...
- ASP.NET MVC3 Dynamically added form fields model binding
Adding new Item to a list of items, inline is a very nice feature you can provide to your user. Thi ...
- Runloop应用实例
AFNetworking AFURLConnectionOperation 这个类是基于 NSURLConnection 构建的,其希望能在后台线程接收 Delegate 回调.为此 AFNetwor ...
- C#:序列化值与解码二进制
1.将对象序列化为二进制值,供WebBrowser传值: private static byte[] PostDataToBytes(Data postData) { JavaScriptSerial ...