Sequence in the Pocket【思维+规律】】的更多相关文章

Sequence in the Pocket 题目链接(点击) DreamGrid has just found an integer sequence  in his right pocket. As DreamGrid is bored, he decides to play with the sequence. He can perform the following operation any number of times (including zero time): select a…
Sequence in the Pocket Time Limit: 1 Second      Memory Limit: 65536 KB DreamGrid has just found an integer sequence  in his right pocket. As DreamGrid is bored, he decides to play with the sequence. He can perform the following operation any number…
传送门 题意: 给出一个序列,你可以将任意一个数移到最前面: 求最少需要移动多少次,可以是此序列变成非递减序列: 思路: 定义 (ai,aj) 为逆序对 ( i < j , ai > aj ), 求出 aj 的最大值,用变量 curMax 存储: 遍历一遍数组,求解 ans: 对于∀ i ∈[1,n] ①如果 ai < curMax , ans++; ②如果 ai == curMax , 那么需要特殊判断: (2.1)如果 ai 之前不曾出现比 curMax 大的数,不需要移动: (2.…
UVa10025 ? 1 ? 2 ? ... ? n = k problem The problem Given the following formula, one can set operators '+' or '-' instead of each '?', in order to obtain a given k? 1 ? 2 ? ... ? n = k For example: to obtain k = 12 , the expression to be used will be:…
Problem Description 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).   Input The input consists of multiple test cases. Each test case…
时间限制:C/C++ 1秒,其他语言2秒 空间限制:C/C++ 262144K,其他语言524288K 64bit IO Format: %lld 题目描述 IG won the S championship and many people are excited, ii and gg are no exception. After watching the game, the two of them also want to play a game. There is now an infin…
思维~ #include<bits/stdc++.h> using namespace std; ; int a[maxn]; int b[maxn]; int N; int main () { int T; scanf ("%d",&T); while (T--) { scanf ("%d",&N); ;i<=N;i++) { scanf ("%d",&a[i]); b[i]=a[i]; } sort…
cf1216E2 Numerical Sequence (hard version) 题目大意 一个无限长的数字序列,其组成为\(1 1 2 1 2 3 1.......1 2 ... n...\),即重复的\(1~1,1~2....1~n\),给你一个k,求第k(k<=1e18)个数字是什么 solution 暴力枚举妥妥T掉 手摸能看出来每一个单独的序列是等差数列 然后每一个序列的数字个数也是等差数列 如果预处理出等差数列的数组,那么询问的时候直接调用即可 参考lzz的blogs 说说这一段…
Description Colossal! — exclaimed Hawk-nose. — A programmer! That's exactly what we are looking for. Arkadi and Boris Strugatsky. Monday starts on Saturday Reading the book "Equations of Mathematical Magic" Roman Oira-Oira and Cristobal Junta fo…
感谢http://blog.csdn.net/black_miracle/article/details/52567718 题意:有一壶水, 体积在 L和 R之间, 有两个杯子, 你要把水倒到两个杯子里面, 使得杯子水体积几乎相同(体积的差值小于等于1), 并且使得壶里剩下水体积不大于1. 你无法测量壶里剩下水的体积, 问最小需要倒水的次数. 题解: r<=1  不用倒就行了  0次 r<=2  倒一次就行了  1单位 1次 r<=l+3  先在左边倒入l/2,在右边倒入l/2+1  2…