这题就是判断是等差数列还是等比数列,然后计算结果mod200907 因为数字比较大10的九次方 所以等比用到了快速幂求模 不懂可以看看算法导论,在大数那里有讲 #include <iostream> #include <stdio.h> #include <stdlib.h> #include <string.h> int modPow(__int64 x,__int64 y) //计算x的y次方求余 快速幂不懂可以百度 { __int64 res=1,a=…

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…

A sequence of numbers Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4384    Accepted Submission(s): 1374 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:http://ac.jobdu.com/problem.php?pid=1442 详解链接:https://github.com/zpfbuaa/JobduInCPlusPlus 参考代码: // // 1442 A sequence of numbers.cpp // Jobdu // // Created by PengFei_Zheng on 15/04/2017. // Copyright © 2017 PengFei_Zheng. All rights reserved. /…

UVA1406 - A Sequence of Numbers(树状数组) 题目链接 题目大意: 给定N个数字.给两种操作:C x: 把这N个数字都加上x. Q x:查询这N个数里面有多少个数字和2^x取且大于0.最后把每一个查询的结果求和作为答案. 解题思路: 查询与2^x取且为1,那么就意味这那个符合要求的数的第x位要是1. 可是这里还有所有加上x的操作,能够用一个变量来记录所有C操作的x的和.有这个加数的话,第x位为1的可能能够通过低位进位得到,所以这里开16个树状数组,第i个树状数组存放…

[九度OJ]题目1442:A sequence of numbers 解题报告 标签(空格分隔): 九度OJ 原题地址:http://ac.jobdu.com/problem.php?pid=1442 题目描述: 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, a…

题目地址:http://ac.jobdu.com/problem.php?pid=1442 题目描述: 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 re…

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…

题目传送门 题意:n个数,两种操作,一是每个数字加x,二是查询& (1 << T) == 1 的个数 分析:因为累加是永远的,所以可以离线处理.树状数组点是c[16][M] 表示数字x%(1 << j) 后的数字pos,考虑第j位的个数.当询问时根据add不同的值不同的处理情况. #include <bits/stdc++.h> using namespace std; typedef long long ll; const int N = 1e5 + 5; c…