题目链接:https://vjudge.net/problem/HDU-4686 Arc of Dream Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 5506    Accepted Submission(s): 1713 Problem Description An Arc of Dream is a curve defined…

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 Problem Description An Arc of Dream is a curve defined by following function: wherea0 = A0ai = a…

Arc of Dream Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 2010    Accepted Submission(s): 643 Problem Description An Arc of Dream is a curve defined by following function:wherea0 = A0ai = ai-…

Arc of Dream Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others) Total Submission(s): 932    Accepted Submission(s): 322 Problem Description An Arc of Dream is a curve defined by following function:where a0 = A0 ai = a…

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others) Total Submission(s): 3354    Accepted Submission(s): 1055 Problem Description An Arc of Dream is a curve defined by following function: where a0 = A0 ai = ai-1*AX+AY…

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+By; 求∑ai*bi(i=0-->n-1). n不超过1018,A0,Ax,Ay,B0,Bx,BY不超过2*109. 题目分析: 因为n很大,不可能用递推来做,这个时候就想到了矩阵的方法.构造了好几个满足要求的,但都是仅仅满足ai或者bi的,最后才发现,把ai*bi按递推式展开, ai*bi=A…

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*BX+BY 求AoD(n) n=0时答案为0!!!! 题解 具体的矩阵构建思路指导可以参考例题链接. 这里仅提供运算过程. Ai=Ai-1*AX+AY Bi=Bi-1*BX+BY AiBi=(Ai-1*AX+AY)(Bi-1*BX+BY)    =AX*BX*Ai-1*Bi-1+AX*BY*Ai-1…

link: http://acm.hdu.edu.cn/showproblem.php?pid=4686 构造出来的矩阵是这样的:根据题目的ai * bi = ……,可以发现 矩阵1 * 矩阵3 = 矩阵2.然后就是矩阵快速幂了. 1 1 ai bi ai*bi Si 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 1 ai+1 bi+1 ai+1*bi+1 Si+1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 1 AY…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4686 因为ai = ai-1*AX+AY ,bi = bi-1*BX+BY ,那么ai*bi=AX*BX*A*ai-1*bi-1+AX*BY*ai-1+BX*AY*bi-1+AY*BYAY.令Sn为ai*bi前n项的和,Sn=Sn-1 + an*bn,因此我们可以构造一个如下的转移矩阵: 然后矩阵乘法优化就可以了... 注意此题n=0的情况! 其实矩阵大小只要5就可以了,那几个常数项可以合并到一列.…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4686 题目意思:给出一个n,算这个式子,给出A0,B0,AX,AY,然后存在以下的递推关系. a0 = A0ai = ai-1*AX+AYb0 = B0bi = bi-1*BX+BY 构造矩阵的思路先算ai,bi,然后算Aod(i),然后再求和,说明这个一定是一个4*4的矩阵,我们可以构造以下系数矩阵(构造矩阵技巧可能会准备专门总结一下): [    1 ,          0 ,    0 ,…