传送门:HDU 5895 Mathematician QSC 这是一篇很好的题解,我想讲的他基本都讲了http://blog.csdn.net/queuelovestack/article/details/52577212 [分析]一开始想简单了,对于a^x mod p这种形式的直接用欧拉定理的数论定理降幂了 结果可想而知,肯定错,因为题目并没有保证gcd(x,s+1)=1,而欧拉定理的数论定理是明确规定的 所以得另谋出路 那么网上提供了一种指数循环节降幂的方法 具体证明可以自行从网上找一找 有…

Mathematician QSC Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Problem Description QSC dream of becoming a mathematician, he believes that everything in this world has a mathematical law. Through unremitting e…

矩阵快速幂,欧拉定理. $g(n)$递推式:$g(n)=5g(n-1)+5g(n-2)-g(n-3)$,可以构造矩阵快速求递$n$项,指数很大,可以利用欧拉定理降幂. #pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include&l…

Mathematician QSC Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 229    Accepted Submission(s): 114 Problem Description QSC dream of becoming a mathematician, he believes that everything in t…

题目链接: Mathematician QSC Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 189    Accepted Submission(s): 90 Problem Description QSC dream of becoming a mathematician, he believes that everything…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5895 f(n)=f(n-2)+2*f(n-1) f(n)*f(n-1)=f(n-2)*f(n-1)+2*f(n-1)*f(n-1); 2*f(n-1)*f(n-1)=f(n)*f(n-1)-f(n-2)*f(n-1); 累加可得 g(n) = f(n)*f(n+1)/2   然后这个公式:A^x % m = A^(x%phi(m)+phi(m)) % m (x >= phi(m))   反正比赛没做…

1001 Resident Evil 1002 List wants to travel 1003 hannnnah_j’s Biological Test 1004 Mathematician QSC 1005 Running King 1006 The Game 1007 odd-even number 1008 oasis in desert 1009 QSC and Master 1010 Count primes 哈哈哈哈...在高铁上打代码哟><…

QSC and Master Problem Description   Every school has some legends, Northeastern University is the same. Enter from the north gate of Northeastern University,You are facing the main building of Northeastern University.Ninety-nine percent of the stude…

题目链接   http://acm.hdu.edu.cn/showproblem.php?pid=5900 题意:给出序列$A_{i}.key$和$A_{i}.value$,若当前相邻的两个数$A_{i}.key$和$A_{i+1}.key$的最大公约数大于1,则可以把这两个数消去,同时消去$A_{i}.value$和$A_{i+1}.value$,每次消去得到的分数为$A_{i}$和$A_{i+1}$的value值,问最大可能得分. 注意:当$A_{i}$和$A_{i+1}被$消去后,$A_{…

题意: 给n件物品,有key和value 每次可以把相邻的 GCD(key[i], key[i+1]) != 1 的两件物品,问移除的物品的总value最多是多少 key : 1 3 4 2  移除34以后12也相邻了,可以移除 分析: 先预处理出所有GCD[l][r], 意味 l <= i <= r的区域可以全部移除, 用记忆化搜索处理 然后 dp[i] 代表到 i 为止可以得到的最大value和 if (G[l][r]) dp[r] = max(dp[r], dp[l-1] + sum[r…