Jerry's trouble Problem's Link:   http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1556 Mean: 略. analyse: 水题,直接快速幂. Time complexity: O(n) Source code:  // Memory Time // 1347K 0MS // by : crazyacking // 2015-03-29-19.18 #include<map> #include<que…

题目链接:http://acm.csu.edu.cn/csuoj/problemset/problem?pid=1556 Description Jerry is caught by Tom. He was penned up in one room with a door, which only can be opened by its code. The code is the answer of the sum of the sequence of number written on th…

[题目链接]:click here [题目大意]:计算x1^m+x2^m+..xn^m(1<=x1<=n)( 1 <= n < 1 000 000, 1 <= m < 1000) [解题思路]:高速幂取模 代码: solution one: #include<bits/stdc++.h> #define LL long long using namespace std; const LL mod=(LL)1e9+7; LL pow_mod(LL a,LL p…

1556: Jerry's trouble Time Limit: 10 Sec  Memory Limit: 256 MBSubmit: 787  Solved: 317[Submit][Status][Web Board] Description Jerry is caught by Tom. He was penned up in one room with a door, which only can be opened by its code. The code is the answ…

题 Description Jerry is caught by Tom. He was penned up in one room with a door, which only can be opened by its code. The code is the answer of the sum of the sequence of number written on the door. The type of the sequence of number is 1^m + 2^m + 3…

题目链接 Problem Description Galen Marek, codenamed Starkiller, was a male Human apprentice of the Sith Lord Darth Vader. A powerful Force-user who lived during the era of the Galactic Empire, Marek originated from the Wookiee home planet of Kashyyyk as…

题目链接 题意: 思路: 直接拿别人的图,自己写太麻烦了~ 然后就可以用矩阵快速幂套模板求递推式啦~ 另外: 这题想不到或者不会矩阵快速幂,根本没法做,还是2013年长沙邀请赛水题,也是2008年Google Codejam Round 1A的C题. #include <bits/stdc++.h> typedef long long ll; const int N = 5; int a, b, n, mod; /* *矩阵快速幂处理线性递推关系f(n)=a1f(n-1)+a2f(n-2)+.…

非010串 基准时间限制:1 秒 空间限制:131072 KB 分值: 80 如果一个01字符串满足不存在010这样的子串,那么称它为非010串. 求长度为n的非010串的个数.(对1e9+7取模)   Input 一个数n,表示长度.(n<1e15) Output 长度为n的非010串的个数.(对1e9+7取模) Input示例 3 Output示例 7 解释: 000 001 011 100 101 110 111 读完题,这样的题目肯定是能找到规律所在的,要不然数据太大根本无法算.假设现在…

题意: 给n(1<n<),求(s1+s2+s3+...+sn)mod(1e9+7).其中si表示n由i个数相加而成的种数,如n=4,则s1=1,s2=3.                         (全题文末) 知识点: 整数n有种和分解方法. 费马小定理:p是质数,若p不能整除a,则 a^(p-1) ≡1(mod p).可利用费马小定理降素数幂. 当m为素数,(m必须是素数才能用费马小定理) a=2时.(a=2只是题中条件,a可以为其他值) mod m =  *      //  k=…

题目 Source http://codeforces.com/contest/632/problem/E Description A thief made his way to a shop. As usual he has his lucky knapsack with him. The knapsack can contain k objects. There are n kinds of products in the shop and an infinite number of pro…