ACM的探索之Just Skip The Problem】的更多相关文章

-----------------心怀虔诚,奋勇前进,fighting!!!!!! Problem Description: inclusively:          包括一切地;包含地 simultaneously:   同时 index number   : 指数; 索引编号 if and only if:       当且仅当 -----Rough Translation: Input: Output: For each test case, output one line contai…
2019 Multi-University Training Contest 2: 1010 Just Skip The Problem 自闭记 题意 多测.每次给你一个数\(n\),你可以同时问无数次,每次问的是一个数\(y_i\),你会得到的回答是\(x\&y_i\)是否为\(y_i\),问你问的最少的次数的种数\(\%1e6+3\),可以调换顺序 10min 得出数学方法 求\(n!\%1e6+3\). 50min 自闭 5min 写出裸暴力,优化10min,交一发TLE 30min 网上…
链接:https://ac.nowcoder.com/acm/contest/554/G Now we have a function f(x): int f ( int x ) {     if ( x == 0 ) return 0;     return f ( x / 10 ) + x % 10; } For a given interval [A, B] (1 <= A <= B <= 10^9), calculate how many integer x that mod f…
Problem Description In Land waterless, water is a very limited resource. People always fight for the biggest source of water. Given a sequence of water sources with a1,a2,a3,...,an representing the size of the water source. Given a set of queries eac…
Description Assuming a finite – radius “ball” which is on an N dimension is cut with a “knife” of N-1 dimension. How many pieces will the “ball” be cut into most?However, it’s impossible to understand the following statement without any explanation.L…
1041: XX's easy problem Time Limit: 1 Sec  Memory Limit: 128 MB Submit: 41  Solved: 7 [Submit][Status][Web Board] Description XX is a good student who likes to ask questions.But sometimes when you get the problem, you are not willing to answer his qu…
http://acm.hdu.edu.cn/showproblem.php?pid=6600 题意:给你一个数x,允许你多次询问yi,然后回答你x xor yi 是否等于yi,询问尽量少的次数以保证能求出xi是几,求出这样询问次数最少的询问方案数. 最优方案就是对每位进行询问,每次只让yi的一位是1,一共n个二进制位.方案数为n! #include<bits/stdc++.h> using namespace std; #define ll long long ; ; ll a[maxn];…
Problem Desciption: 百度翻译后的汉化: 参见博客:https://www.cnblogs.com/zxcoder/p/11253099.html https://blog.csdn.net/Cassie_zkq/article/details/97255683https://blog.csdn.net/Cassie_zkq/article/details/97255683…
[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 123...n中末尾0的个数 只会由素因子2和5的个数决定且等于 Min{cnt[2],cnt[5]} 且素因子2的个数一定会比5多; 所以n!的末尾0的个数 等于for (int i = 1;i <= n;i++)中所有i的5因子的个数和 枚举一下就好 [代码] #include <bits/stdc++.h> using namespace std; const int N = 1e7; int n; vector&…
题意:询问n!模1e6+7的结果 n<=1e9 思路: #include<bits/stdc++.h> using namespace std; typedef long long ll; typedef unsigned int uint; typedef unsigned long long ull; typedef pair<int,int> PII; typedef pair<ll,ll> Pll; typedef vector<int> VI…