hdu 2817 A sequence of numbers(快速幂)】的更多相关文章

Problem Description Xinlv wrote some sequences on the paper a long time ago, they might be arithmetic or geometric sequences. The numbers are not very clear now, and only the first three numbers of each sequence are recognizable. Xinlv wants to know…
A sequence of numbers Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4550    Accepted Submission(s): 1444 Problem Description Xinlv wrote some sequences on the paper a long time ago, they might…
http://acm.hdu.edu.cn/showproblem.php?pid=2817 __int64 pow_mod (__int64 a, __int64 n, __int64 m)快速幂取模函数. A sequence of numbers Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4047    Accepted Su…
题目链接:Recursive sequence 题意:给出前两项和递推式,求第n项的值. 题解:递推式为:$F[i]=F[i-1]+2*f[i-2]+i^4$ 主要问题是$i^4$处理,容易想到用矩阵快速幂,那么$i^4$就需要从$(i-1)$转移过来. $ i^4 = (i-1)^4 + 4*(i-1)^3 + 6*(i-1)^2 + 4*(i-1) + 1$ $f_i$ $f_{i-1}$ $i^4$ $i^3$ $i^2$ $i$ $1$ = $f_{i-1}$ $f_{i-2}$ $(i…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1005 题意: 数列{f(n)}: f(1) = 1, f(2) = 1, f(n) = ( A*f(n-1) + B*f(n-2) ) MOD 7 给定A.B.n,求f(n). (1<=n<=100,000,000) 题解: 大水题~ (*/ω\*) 矩阵快速幂. 初始矩阵start: 特殊矩阵special: 所求矩阵ans: ans = start * special^(n-1) ans的第一…
题目 第一次做是看了大牛的找规律结果,如下: //显然我看了答案,循环节点是48,但是为什么是48,据说是高手打表出来的 #include<stdio.h> int main() { ],a,b,i,n; f[]=;f[]=; while(scanf("%d%d%d",&a,&b,&n)!=EOF) { &&b==&&n==)break; ;i<;i++) { f[i]=(a*f[i-])%+(b*f[i-])%…
思路:一开始不会n^4的推导,原来是要找n和n-1的关系,这道题的MOD是long long 的,矩阵具体如下所示 最近自己总是很坑啊,代码都瞎吉坝写,一个long long的输入写成%d一直判我TLE,一度怀疑矩阵快速幂地复杂度orz 代码: #include<set> #include<cstring> #include<cstdio> #include<algorithm> #define ll long long const int maxn = 7…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2817 题目大意:给出三个数,来判断是等差还是等比数列,再输入一个n,来计算第n个数的值. #include <iostream> #include <cstdio> #include <cmath> #define m 200907 using namespace std; __int64 fun(__int64 j,__int64 k) { __int64 s=; whi…
A number sequence is defined as follows: f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7. Given A, B, and n, you are to calculate the value of f(n). InputThe input consists of multiple test cases. Each test case contains 3 integers A,…
Sequence  Accepts: 59  Submissions: 650  Time Limit: 2000/1000 MS (Java/Others)  Memory Limit: 65536/65536 K (Java/Others) Problem Description \ \ \ \    Holion August will eat every thing he has found. \ \ \ \    Now there are many foods,but he does…