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A square-free integer is an integer which is indivisible by any square number except 11. For example, 6 = 2 \cdot 36=2⋅3 is square-free, but 12 = 2^2 \cdot 312=22⋅3 is not, because 2^222 is a square number. Some integers could be decomposed into prod…
ACM-ICPC 2018 南京赛区网络预赛 E题 题目链接: https://nanti.jisuanke.com/t/30994 Dlsj is competing in a contest with n (0 < n \le 20)n(0<n≤20) problems. And he knows the answer of all of these problems. However, he can submit ii-th problem if and only if he has s…
题目链接:https://nanti.jisuanke.com/t/30991 Feeling hungry, a cute hamster decides to order some take-away food (like fried chicken for only 3030 Yuan). However, his owner CXY thinks that take-away food is unhealthy and expensive. So she demands her hams…
J. Sum 26.87% 1000ms 512000K   A square-free integer is an integer which is indivisible by any square number except 11. For example, 6 = 2 \cdot 36=2⋅3 is square-free, but 12 = 2^2 \cdot 312=22⋅3 is not, because 2^222 is a square number. Some integer…
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A. An Olympian Math Problem 54.28% 1000ms 65536K   Alice, a student of grade 66, is thinking about an Olympian Math problem, but she feels so despair that she cries. And her classmate, Bob, has no idea about the problem. Thus he wants you to help him…
题目链接:https://nanti.jisuanke.com/t/30991 2000ms 262144K   Feeling hungry, a cute hamster decides to order some take-away food (like fried chicken for only 3030 Yuan). However, his owner CXY thinks that take-away food is unhealthy and expensive. So she…
轻轻松松也能拿到区域赛名额,CCPC真的好难 An Olympian Math Problem 问答 只看题面 54.76% 1000ms 65536K   Alice, a student of grade 66, is thinking about an Olympian Math problem, but she feels so despair that she cries. And her classmate, Bob, has no idea about the problem. T…
262144K   There are NN cities in the country, and MM directional roads from uu to v(1\le u, v\le n)v(1≤u,v≤n). Every road has a distance c_ici​. Haze is a Magical Girl that lives in City 11, she can choose no more than KK roads and make their distanc…
ACM-ICPC 2018 南京赛区网络预赛 A. An Olympian Math Problem 计算\(\sum_{i=1}^{n-1}i\cdot i!(MOD\ n)\) \(\sum_{i=1}^{n-1}i\cdot i! = \sum_{i=1}^{n-1}[(i+1)!-i!](MOD\ n)=n!-1!(MOD\ n)=n-1\) #include<bits/stdc++.h> using namespace std; int T; long long n; int mai…