HDU 5063 Operation the Sequence 题目链接 把操作存下来.因为仅仅有50个操作,所以每次把操作逆回去执行一遍,就能求出在原来的数列中的位置.输出就可以 代码: #include <cstdio> #include <cstring> #include <algorithm> using namespace std; typedef long long ll; const int N = 100005; const ll MOD = 1000…

Operation the Sequence                                                                     Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)                                                                            …

题目链接:pid=5063" target="_blank">http://acm.hdu.edu.cn/showproblem.php?pid=5063 Problem Description You have an array consisting of n integers: a1=1,a2=2,a3=3,-,an=n. Then give you m operators, you should process all the operators in order…

http://acm.hdu.edu.cn/showproblem.php?pid=5063 题目大意: 题目意思还是比较简单.所以就不多少了.注意这句话,对解题有帮助. Type4: Q i query current value of a[i], this operator will have at most 50. 解题思路: 因为给定n和m都是100000,我们每一步都做具体的操作,时间将是O(n*m),肯定超时.最开始的时候 我怎么想都不知道怎么解决.以为是线段树.比赛完了以后,看了解…

http://acm.hdu.edu.cn/showproblem.php?pid=5063 思路:因为3查询最多50,所以可以在查询的时候逆操作找到原来的位置,然后再求查询的值. #include <cstdio> #include <iostream> #include <cstring> #include <algorithm> #define ll long long using namespace std; ; int n,m; ]; ]; ll…

注意到查询次数不超过50次,那么能够从查询位置逆回去操作,就能够发现它在最初序列的位置,再逆回去就可以求得当前查询的值,对于一组数据复杂度约为O(50*n). Operation the Sequence Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 463    Accepted Submission(s): 187 Problem…

题目链接: hdu:http://acm.hdu.edu.cn/showproblem.php?pid=5273 bc:http://bestcoder.hdu.edu.cn/contests/contest_chineseproblem.php?cid=604&pid=1002 Dylans loves sequence  Accepts: 250  Submissions: 806  Time Limit: 2000/1000 MS (Java/Others)  Memory Limit:…

#include <cstdio> #include <cstring> #include <cstdlib> #define MAXN 100005 #define MOD 1000000007 , rear = ; int n; void query(int index) { int r = rear; ; )>>; while (r-- > front) { ) { ++square; } ) { index = n+-index; } else…

Problem Description You have an array consisting of n integers: a1=1,a2=2,a3=3,-,an=n. Then give you m operators, you should process all the operators in order. Each operator is one of four types: Type1: O 1 call fun1(); Type2: O 2 call fun2(); Type3…

题意 两个1e5的数组a,b,定义\(S(t)=\left \lfloor \frac{t-b_i}{a_i} \right \rfloor\),有三个操作 1 x y:将\(a[x]\)变为\(y\) 2 x y:将\(b[x]\)变为\(y\) 3 x:求使得\(S(t)\geq k\)的最小\(k\) 其中\(a_i\leq 1000\),\(b_i,k\leq 1e9\) 思路 这题主要突破口在于\(a_i\leq 1000\) 我们先从下整除下手, \(\left \lfloor \f…