构造矩阵如下: Ai*bi AX*BX AX*BY AY*BX AY*BY 0 a(i-1)*b(i-1) Ai 0 AX 0 AY 0 a(i-1) Bi 0 0 BX BY 0 b(i-1) 1 0 0 0 1 0 1 Sum(i) AX*BX AX*BY AY*BX AY*BY 1 sum(i-1) Sum(i) 表示i项和,sum(i)=sum(i-1)+ai*bi; 求第n次的结果,直接对矩阵作n-1次,利用矩阵快速幂,时间复杂度为10*logn~logn 注意取模爆范围和对n=0特判…

Arc of Dream Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 0    Accepted Submission(s): 0 Problem Description An Arc of Dream is a curve defined by following function:wherea0 = A0ai = ai-1*AX+…

Arc of Dream Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 419    Accepted Submission(s): 148 Problem Description An Arc of Dream is a curve defined by following function:wherea0 = A0ai = ai-1…

Problem Description An Arc of Dream is a curve defined by following function: where a0 = A0 ai = ai-*AX+AY b0 = B0 bi = bi-*BX+BY What ,,,?   Input There are multiple test cases. Process to the End of File. Each test nonnegative integers as follows:…

标题效果:你就是给你一程了两个递推公式公式,第一个让你找到n结果项目. 注意需要占用该公式的复发和再构造矩阵. Arc of Dream Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others) Total Submission(s): 2092    Accepted Submission(s): 664 Problem Description An Arc of Dream is a…

Arc of Dream [题目链接]Arc of Dream [题目类型]矩阵 &题解: 这题你做的复杂与否很大取决于你建的矩阵是什么样的,膜一发kuangbin大神的矩阵: 还有几个坑点:当n是0 输出0;建矩阵时是相乘的一定要取模M,因为如果不取模最大的情况是1e9*2e9*2e9,爆long long 这块坑了我好长时间. &代码: #include <cstdio> #include <bitset> #include <iostream> #…

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: 65535/65535 K (Java/Others) Total Submission(s): 2164    Accepted Submission(s): 680 Problem Description An Arc of Dream is a curve defined by following fun…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4686 题意: 思路: #include <iostream>#include <cstdio>#include <string.h>#include <algorithm>#include <cmath>#include <vector>#include <queue>#include <set>#include…

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 最后的结果mod 1,000,000,007 n<=10^18. 分析:ai*bi=(ai-1 *ax+ay)*(bi-1 *bx+by) =(ai-1 * bi-1 *ax*bx)+(ai-1 *ax*by)+(bi-1 *bx*ay)+(ay*by) 设p=ax*bx,  q=ax*by, …

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的递推式题目已经给出, Ai*Bi=Ax*Bx*(Ai-1*Bi-1)+Ax*By*Ai-1+Bx*Ay*Bi-1+Ay*By AoD(n)=AoD(n-1)+AiBi 构造向量I{AoD(i-1),Ai*Bi,Ai,Bi,1} 初始向量为I0={0,A0*B0,A0,B0,1} 构造矩阵A{ 1,0,0,0,…