The Android University ACM Team Selection Contest Time Limit: 1000ms   Memory limit: 65536K  有疑问?点这里^_^ 题目描述  Now it's 20000 A.D., and the androids also participate in the ACM Inter-national Collegiate Programming Contest (ACM/ICPC). In order to sele…

Identifiers Time Limit: 1000ms   Memory limit: 65536K  有疑问?点这里^_^ 题目描述  Identifier is an important concept in the C programming language. Identifiers provide names for several language elements, such as functions, variables, labels, etc. An identifie…

Crack Mathmen Time Limit: 1000ms   Memory limit: 65536K  有疑问?点这里^_^ 题目描述  Since mathmen take security very seriously, they communicate in encrypted messages. They cipher their texts in this way: for every characther c in the message, they replace c w…

Ivan comes again! Time Limit: 1000ms   Memory limit: 65536K  有疑问?点这里^_^ 题目描述 The Fairy Ivan gave Saya three problems to solve (Problem F). After Saya finished the first problem (Problem H), here comes the second.This is the enhanced version of Proble…

Clockwise Time Limit: 1000ms   Memory limit: 65536K  有疑问?点这里^_^ 题目描述 Saya have a long necklace with N beads, and she signs the beads from 1 to N. Then she fixes them to the wall to show N-1 vectors – vector i starts from bead i and end up with bead i…

Balloons Time Limit: 1000MS Memory limit: 65536K 题目描述 Both Saya and Kudo like balloons. One day, they heard that in the central park, there will be thousands of people fly balloons to pattern a big image.They were very interested about this event, an…

Shopping Time Limit: 1000MS Memory limit: 65536K 题目描述 Saya and Kudo go shopping together.You can assume the street as a straight line, while the shops are some points on the line.They park their car at the leftmost shop, visit all the shops from left…

Hello World! Time Limit: 1000MS Memory limit: 65536K 题目描述 We know that Ivan gives Saya three problems to solve (Problem F), and this is the first problem.“We need a programmer to help us for some projects. If you show us that you or one of your frien…

A. Hacker Cups and Balls 二分答案,将$\geq mid$的数看成$1$,$<mid$的数看成$0$,用线段树进行区间排序检查即可.时间复杂度$O(n\log^2n)$. #include<cstdio> #include<algorithm> using namespace std; const int N=100010,M=262150; int n,m,i,a[N],e[N][2],l,r,MID,ans; int len[M],c1[M],ta…

  D 考虑每个点被删除时其他点对它的贡献,然后发现要求出距离为1~k的点对有多少个. 树分治+FFT.分治时把所有点放一起做一遍FFT,然后减去把每棵子树单独做FFT求出来的值. 复杂度$nlog^2n$ #include<bits/stdc++.h> #define N 270000 #define pi acos(-1) #define ll long long #define inf 0x3f3f3f3f using namespace std; const int p = 10000…