点击此处即可传送 hdu 3509题目大意:F1 = f1, F2 = f2;;
F(n) = a*F(n-1) + b*F(n-2);
S(n) = F1^k + F2^k +….+Fn^k;
求S(n) mod m;
解题思路:
1:首先一个难题就是怎么判断矩阵的维数(矩阵的维数是个变量)
解决方法:开一个比较大的数组,然后再用一个公有变量记一下就行了,具体详见代码;
2:k次方,找规律;
具体上代码吧:/*
2015 - 8 - 16 晚上
Author: ITAK今日的我要超越昨日的我,明日的我要胜过今日的我,
以...
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2132题意很简单。这里采用O(1)算法。 1 #include <cstdio>2 3int main()4{5int n,ans;6while(~scanf("%d",&n))7 {8int m,cnt=0;9for(int i=0;i<n;i++)
10 {
11 scanf("%d",&m);
12if(cnt==0) ans=m,cnt++;
13elseif(ans==m) cnt++;
14else cnt--;
15 }
16 printf("%d\n",ans);
17 }
18return0;
19 }View Code 原文:ht...
input::-webkit-outer-spin-button,
input::-webkit-inner-spin-button {-webkit-appearance: none;
}
input[type="number"]{-moz-appearance: textfield;
} 原文:https://www.cnblogs.com/shui1993/p/10855989.html
Given an array containing n distinct numbers taken from 0,1, 2, ..., n, find the one that is missing from the array.
For example,
Given nums = [0, 1, 3] return 2.Note:
Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?Credits:Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.第一次AC代码...
Given an array of integers, every element
appears twice except for one. Find that single one.Could you implement it without using extra memory?Ref: http://www.cnblogs.com/feiling/p/3349654.html[解题思路]以前看书时遇到过,一个数组中有一个元素出现一次,其他每个元素都出现两次要求空间复杂度为O(1)由于两个相同的元素异或的结果为0,而0^a==a,将数组中所有元素都进行异或,得到结果就是只出现一次的元素 publicc...
Total jobs = 1Launching Job 1 out of 1Number of reduce tasks determined at compile time: 1In order to change the average load for a reducer (in bytes): set hive.exec.reducers.bytes.per.reducer=<number>In order to limit the maximum number of reducers: set hive.exec.reducers.max=<number>In order to set a constant number of reducers: set mapreduce.job.reduces=<number>Interrupting... Be patient, th...
Given two strings representing two complex numbers.You need to return a string representing their multiplication. Note i2 = -1 according to the definition.Example 1:Input: "1+1i", "1+1i"
Output: "0+2i"
Explanation: (1 + i) * (1 + i) = 1 + i2 + 2 * i = 2i, and you need convert it to the form of 0+2i.
Example 2:Input: "1+-1i", "1+-1i"
Output: "0+-2i"
Explanation: (1 - i) * (1 - i) = 1 + i2 - 2 * i ...
17. Letter Combinations of a Phone Number Problem‘s Link ----------------------------------------------------------------------------
Mean: 给你一个数字串,输出其在手机九宫格键盘上的所有可能组合.analyse:使用进位思想来枚举所有组合即可.Time complexity: O(N) view code/** * ----------------------------------------------------------------- * Copyright (c) 2016 crazyacking.All rights reserved. * ------...
Given a string text, you want to use the characters of text to form as many instances of the word "balloon" as possible.You can use each character in text at most once. Return the maximum number of instances that can be formed.Example 1:Input: text = "nlaebolko"
Output: 1
Example 2:Input: text = "loonbalxballpoon"
Output: 2
Example 3:Input: text = "leetcode"
Output: 0Constraints:1 <= text.length <...
这道题可以看作是26进制问题。26进制就是最多可以用26个不同的符号组合在一起来表示一个数值。因此,ABC代表的数值就是:A*262+B*261+C*260=1*262+2*261+3*260,其他字母组合以此类推。下面这种方法是一种很直观的思路,但是要调用C++库函数pow(x,a),计算时间会变长。pow(x,a)可以求得x的a次幂,‘A‘的ASCII码是64。class Solution {
public:int titleToNumber(string s) {int n=s.size()-1;int result=0;for(int i=0;i<=n;i++){...
Question:Given an array of integers, every element appears three times except for one. Find that single one.Note:Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?Solution: 1class Solution {2public:3int singleNumber(vector<int>& nums) {4 sort(nums.begin(),nums.end());5for(int i=0;i<nums.size();i+=3)6 {7if(nums[i]!=nums[i+1])8 {9r...
HDU-1005题解:每次的数都会对7进行mod操作,mod之后数的范围一定在[0,6]之间,如果每次都循环到n的话明显会TLE,那么就说明这个结果会出现一个周期,我们只要找到这个周期,然后根据周期进行mod,然后找到结果就好了。PS:讨论区的最简代码的mod48操作我也不是很理解,但是有人是说错的,hdu数据水而已,2333。(反正我觉得找周期靠谱)代码: 1 #include<iostream>2usingnamespace std;3constint N = 7;4longlong A, B, n,f1,f2,f...
tabs as
(
select ROW_NUMBER() over(partition by customerID order by totalPrice) as rows,customerID,totalPrice, DID from OP_Order )
select MAX(rows) as ‘下单次数‘,customerID from tabs group by customerID with tabs as (
select ROW_NUMBER() over(partition by customerID order by insDT) as rows,customerID,totalPrice, DID from OP_Order
) select * from tabs
where totalPrice in
(
s...
Source insight最新版本3572下载链接:http://www.sourceinsight.com/down35.html, http://www.sourceinsight.com/distribute/Si3570Setup.exe注冊号:SI3US-205035-364481.卸载掉旧版安装新版, 可能会有提示" An invalid source insight serial number was detected",然后不能使用.解决方法:(1) 快捷键:WIN+R,或者 開始 -> 附件 -> 执行,输入regedit,打开注冊表编辑器(2) 找到source insight 的安装路径32位系统路径例如以下:64位...
Given a digit string, return all possible letter combinations that the number could represent.
A mapping of digit to letters (just like on the telephone buttons) is given below.Input:Digit string "23"
Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].
Note:
Although the above answer is in lexicographical order, your answer could be in any order you want.
采用两种方式都可以实现DFS和BFS...