题目 A Simple Math Problem 解析 矩阵快速幂模板题 构造矩阵 \[\begin{bmatrix}a_0&a_1&a_2&a_3&a_4&a_5&a_6&a_7&a_8&a_9\\ 1&0&0&0&0&0&0&0&0&0\\ 0&1&0&0&0&0&0&0&0&0\\ 0&…

A Simple Math Problem [题目链接]A Simple Math Problem [题目类型]矩阵快速幂 &题解: 这是一个模板题,也算是入门了吧. 推荐一个博客:点这里 跟着这个刷,应该就可以了 &代码: #include <cstdio> #include <iostream> #include <set> #include <cmath> #include <cstring> #include <al…

题意:略 简单的矩阵快速幂就行了 #include <iostream> #include <cstdio> #include <cstring> using namespace std; #define LL long long #define N 10 int m; struct node{ int mat[N][N]; node operator *(const node &x){ node tmp; memset(tmp.mat,0,sizeof(tmp…

任意门:http://acm.hdu.edu.cn/showproblem.php?pid=1757 A Simple Math Problem Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6621    Accepted Submission(s): 4071 Problem Description Lele now is thin…

A Simple Math Problem Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2441    Accepted Submission(s): 1415 Problem Description Lele now is thinking about a simple function f(x). If x < 10 f(x) =…

A Simple Math Problem Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1697    Accepted Submission(s): 959 Problem Description Lele now is thinking about a simple function f(x). If x < 10 f(x) =…

A Simple Math Problem Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4331    Accepted Submission(s): 2603 Problem Description Lele now is thinking about a simple function f(x).If x < 10 f(x) =…

题目地址:HDU 1757 最终会构造矩阵了.事实上也不难,仅仅怪自己笨..= =! f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + -- + a9 * f(x-10) 构造的矩阵是:(我代码中构造的矩阵跟这个正好是上下颠倒过来了) |0 1 0 ......... 0|    |f0|   |f1 | |0 0 1 0 ...... 0|    |f1|   |f2 | |...................1| *  |..| = |...…

题目链接 题意 :给你m和k, 让你求f(k)%m.如果k<10,f(k) = k,否则 f(k) = a0 * f(k-1) + a1 * f(k-2) + a2 * f(k-3) + …… + a9 * f(k-10);思路 :先具体介绍一下矩阵快速幂吧,刚好刚刚整理了网上的资料.可以先了解一下这个是干嘛的,怎么用. 这个怎么弄出来的我就不说了,直接看链接吧,这实在不是我强项,点这儿,这儿也行 //HDU 1757 #include <iostream> #include <s…

Description Lele now is thinking about a simple function f(x). If x < 10 f(x) = x. If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10); And ai(0<=i<=9) can only be 0 or 1 . Now, I will give a0 ~ a9 and two positive in…

题目 也是和LightOJ 1096 和LightOJ 1065 差不多的简单题目. #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; int num,mod; struct matrix { ][]; }; matrix multiply(matrix x,matrix y)//矩阵乘法 { matrix temp; ;i<num;i++) { ;j<…

题意:有一个递推式f(x) 当 x < 10    f(x) = x.当 x >= 10  f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + -- + a9 * f(x-10) 同时ai(0<=i<=9) 不是 0 就是 1: 现在给你 ai 的数字,以及k和mod,请你算出 f(x)%mod 的结果是多少 思路:线性递推关系是组合计数中常用的一种递推关系,如果直接利用递推式,需要很长的时间才能计算得出,时间无法承受,但是现在我们已知…

<font color = red , size = '4'>下列图表转载自 efreet 链接:传送门 题意:给出递推关系,求 f(k) % m 的值, 思路: 因为 k<2 * 10^9 , m < 10^5,O(n)算法应该是T掉了,当 k >= 10 时 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + -- + a9 * f(x-10),可以理解为这是两个行列是乘积的值,经下面分析可知用矩阵快速幂可搞 下列三个表分别命…

题意:当x < 10时, f(x) = x: 当x >= 10 时,f(x) = a0 * f(x-1) + a1 * f(x-2) +  + a2 * f(x-3) + …… + a9 * f(x-10): ai(0<=i<=9) 只能是0或者1 ,给出a0 ~ a9,k和m,计算f(k)%m(k<2*10^9 , m < 10^5). 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1757 ——>>构造矩阵…

和这一题构造的矩阵的方法同样. 须要注意的是.题目中a0~a9 与矩阵相乘的顺序. #include <iostream> #include <cstdio> #include <algorithm> #include <cmath> #include <cstring> #define N 10 using namespace std; int mod; typedef long long LL; struct matrix { LL a[10…

Problem Description Lele now is thinking about a simple function f(x). If x < f(x) = x. If x >= f(x) = a0 * f(x-) + a1 * f(x-) + a2 * f(x-) + …… + a9 * f(x-); And ai(<=i<=) can only be or . Now, I will give a0 ~ a9 and two positive integers k…

题意:题意很简单,不多说了. 思路: |f(10) |       |a0 a1 a2 ...a8 a9|    |f(9)|| f(9)  |       | 1   0   0 ... 0    0 |    |f(8)|| .....  |   =  | ..    ...    ...   ...    |     | ..   || f(2) |        | 0   0   0 ... 0    0|     |f(1)|| f(1) | | 0   0   0 ... 1  …

Lele now is thinking about a simple function f(x).  If x < 10 f(x) = x.  If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + -- + a9 * f(x-10);  And ai(0<=i<=9) can only be 0 or 1 .  Now, I will give a0 ~ a9 and two positive integers k…

A Simple Math Problem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1645    Accepted Submission(s): 468 Problem Description Given two positive integers a and b,find suitable X and Y to meet th…

A Simple Math Problem Lele now is thinking about a simple function f(x).If x < 10 f(x) = x.If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + -- + a9 * f(x-10);And ai(0<=i<=9) can only be 0 or 1 .Now, I will give a0 ~ a9 and two posit…

CRB and Puzzle Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 483    Accepted Submission(s): 198 Problem Description CRB is now playing Jigsaw Puzzle. There are  kinds of pieces with infinite…

HDU 4965 Fast Matrix Calculation 题目链接 矩阵相乘为AxBxAxB...乘nn次.能够变成Ax(BxAxBxA...)xB,中间乘n n - 1次,这样中间的矩阵一个仅仅有6x6.就能够用矩阵高速幂搞了 代码: #include <cstdio> #include <cstring> const int N = 1005; const int M = 10; int n, m; int A[N][M], B[M][N], C[M][M], CC[N…

题链:http://lightoj.com/volume_showproblem.php?problem=1070 1070 - Algebraic Problem PDF (English) Statistics Forum Time Limit: 2 second(s) Memory Limit: 32 MB Given the value of a+b and ab you will have to find the value of an+bn. a and b not necessar…

题意:一种彩票共同拥有 N 个号码,每注包括 M 个号码,假设开出来的 M 个号码中与自己买的注有 R 个以上的同样号码,则中二等奖,问要保证中二等奖至少要买多少注(1<=R<=M<=N<=8). 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4979 -->>覆盖问题,yy可知是可反复覆盖问题,于是,DLX 上场.. N个 选 R 个,共同拥有 C[N][R] 种选法,每种选法须要被覆盖,相应于 DLX 中的列.. N个…

Problem Description Given two positive integers a and b,find suitable X and Y to meet the conditions:                                                        X+Y=a                                              Least Common Multiple (X, Y) =b   InputInp…

http://acm.hdu.edu.cn/showproblem.php?pid=5974 遇到数学题真的跪.. 题目要求 X + Y = a lcm(X, Y) = b 设c = gcd(x, y); 那么可以表达出x和y了,就是x = i * c; y = j * c; 其中i和j是互质的. 所以lcm(x, y) = i * j * c = b 那么就得到两个方程了. i * c + j * c = a; i * j * c = b; 但是有一个c,三个未知数. 因为i和j互质,所以(i…

Problem Description Given two positive integers a and b,find suitable X and Y to meet the conditions: X+Y=a Least Common Multiple (X, Y) =b   Input Input includes multiple sets of test data.Each test data occupies one line,including two positive inte…

Description Dragon loves lottery, he will try his luck every week. One day, the lottery company brings out a new form of lottery called accumulated lottery. In a normal lottery, you pick 7 numbers from N numbers. You will get reward according to how…

题意:给定a和b,求一组满足x+y=a && lcm(x, y)=b. 析:x+y = a, lcm(x, y) = b,=>x + y = a, x * y = b * k,其中 k = gcd(x, y). 然后第一个式子同时除以k,第二个式子同时除以k*k,那么x/k,和y/k是互质的,那么a/k和b/k也是互质的.所以问题就转化成了 x' + y' = a',x' * y' = b'.然后解方程并判断解的存在即可. 代码如下: #pragma comment(linker,…

题目链接 题意 现有\[x+y=a\\lcm(x,y)=b\]找出满足条件的正整数\(x,y\). \(a\leq 2e5,b\leq 1e9,数据组数12W\). 思路 结论 \(gcd(x,y)=gcd((x+y),lcm(x,y))\) 证明 先证\(gcd(x,y)|gcd((x+y),lcm(x,y))\) 不妨设\(gcd(x,y)=k\),则有\(k\mid x,k\mid y\),则有\(k\mid (x+y)\) -① 又\(k\mid x,x\mid lcm(x,y)\),所…