杭州现场赛的题.BFS+DFS #include <iostream> #include<cstdio> #include<cstring> #define inf 9999999 using namespace std; char mp[105][105]; int sq[5][5]; int step[4][2]={{0,1},{1,0},{0,-1},{-1,0}}; struct pos { int x,y; }; int n,m,prn,x,y,tmp,ans…

Zhuge Liang's Mines Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 239    Accepted Submission(s): 110 Problem Description In the ancient three kingdom period, Zhuge Liang was the most famous an…

Caocao's Bridges Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 194    Accepted Submission(s): 89 Problem Description Caocao was defeated by Zhuge Liang and Zhou Yu in the battle of Chibi. But…

Mex Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 623    Accepted Submission(s): 209 Problem Description Mex is a function on a set of integers, which is universally used for impartial game t…

Save Labman No.004 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 624    Accepted Submission(s): 154 Problem Description Due to the preeminent research conducted by Dr. Kyouma, human beings hav…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4816 2013长春区域赛的D题. 很简单的几何题,就是给了一条折线. 然后一个矩形窗去截取一部分,求最大面积. 现场跪在这题,最后时刻TLE到死,用的每一小段去三分,时间复杂度是O(n log n) , 感觉数据也不至于超时. 卧槽!!!!代码拷回来,今天在HDU一交,一模一样的代码AC了,加输入外挂6s多,不加也8s多,都可AC,呵呵·····(估计HDU时限放宽了!!!) 现场赛卡三分太SXBK…

第一年参加现场赛,比赛的时候就A了这一道,基本全场都A的签到题竟然A不出来,结果题目重现的时候1A,好受打击 ORZ..... 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4800 题目大意:给定C(3,N)支队伍之间对战的获胜概率,再给定一个序列存放队伍编号,每次获胜之后可以选择和当前战胜的对手换队伍.问按给定序列依次挑战全部胜利的最大概率. 解题思路:状压DP dp[i][j]表示使用队伍i从编号j开始挑战全胜的概率,ai[i]表示i位置的队…

#include<iostream> #include<cstring> #include<algorithm> #include<cmath> #include<cstdio> #define Maxn 100 using namespace std; <<],num[Maxn],bag[Maxn][Maxn],g,b,s,now[]; bool vi[Maxn]; int dfs(int S,int now[]) { if(vi[…

The Donkey of Gui Zhou Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 389    Accepted Submission(s): 153 Problem Description There was no donkey in the province of Gui Zhou, China. A trouble m…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4821 字符串题. 现场使用字符串HASH乱搞的. 枚举开头! #include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <queue> #include <map> #include <set> #inclu…

Save Labman No.004 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 979    Accepted Submission(s): 306 Problem Description Due to the preeminent research conducted by Dr. Kyouma, human beings ha…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4815 简单的DP题. #include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <queue> #include <map> #include <set> #include <vector> #i…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4814 进制转换. 现场根据题目给的两个公式,不断更新!!! 胡搞就可以了. 现场3A,我艹,一次循环开大TLE,一次开小WA,太逗了,呵呵 现场源码: #include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <queue> #in…

比赛的时候就预感到这题能出,但是会耗时比较多.结果最后是出了,但是有更简单的题没出. 是不是错误的决策呢?谁知道呢 题目意思: 定义f(x) = x分解质因数出来的因子个数 如 x = p0 * p0 * p0 * p1 * p2,则f(x) = 5 特殊的, f(1) = 0 求 i = [1..n], j = [1..m] 组成的n*m组(i, j)对中,有多少组f( gcd(i,j) ) <= p 考虑简化版本,p = 0,即求有多少组 gcd(i,j) == 1. 见HDU 1695 h…

链接:http://acm.hdu.edu.cn/showproblem.php?pid=4815 [题意] n个题目,每题有各自的分数,A有50%的概率答对一道题目得到相应分数,B想要在至少P的概率上总分不低于A,问B至少要得到多少分. [分析] 最简单粗暴的做法是算出每个可能得到的总分的概率,原问题可以转化成在概率和<=P下A所有可能得到的总分集合中最大分数的最小值为多少,于是答案就是按分数排序后前k项的概率和刚好>=P. 但是计算所有可能的概率的复杂度是O(2n),不能满足我们的需求.细…

/** 题意: 有两种塔,重塔,轻塔.每种塔,能攻击他所在的一行和他所在的一列, 轻塔不 能被攻击,而重塔可以被至多一个塔攻击,也就是说重塔只能被重塔攻击.在一个n*m 的矩阵中,最少放一个塔,可放多个 问,给定p个重塔,q个轻塔,问有多少种放法.. 思路: 1. 一行中有两个重塔, 2. 一列中有两个重塔 3. 在该行及在该行塔所在的列只有一个塔,重塔或者轻塔. 对以上三种情况 挨个处理: 1. 设有i行有两个重塔,j列有两个重塔,则一共占 i+2*j 行, j+2*i列,共用2*(i+j)个…

把一个字符串分成N个字符串 每个字符串长度为m Sample Input12 5 // n mklmbbileay Sample Outputklmbbileay # include <iostream> # include <cstdio> # include <cstring> # include <algorithm> # include <string> # include <cmath> # include <queu…

100MB=10^5KB=10^8B 100MB=100*2^10KB=100*2^20B Sample Input2100[MB]1[B] Sample OutputCase #1: 4.63%Case #2: 0.00% # include <iostream> # include <cstdio> # include <cstring> # include <algorithm> # include <string> # include &…

题目意思: 给定n, expect, a, b 要求你构造一组array[],存放一个1..n的排列,使的下面的程序能输出YES 题目所示代码: bool less_than(x, y) { T++; return x < y; } void work(array[], l, r) { if (l >= r) return; swap(array[(l * A + r * B) / (A + B)], array[r]); int index = l; for (i = l; i < r…

Two Rabbits Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 505    Accepted Submission(s): 260 Problem Description Long long ago, there lived two rabbits Tom and Jerry in the forest. On a sunny…

2013年山东省赛F题 Mountain Subsequences先说n^2做法,从第1个,(假设当前是第i个)到第i-1个位置上哪些比第i位的小,那也就意味着a[i]可以接在它后面,f1[i]表示从第一个开始,以a[i]为结尾的不同递增序列的个数,要加上1,算上本身.正反各跑一遍,答案加一下(f1[i]-1)*(f2[i]-1)优化就是,比a[i]小的,只有a[i]-1个 #include<iostream> #include<cstdio> #include<queue&…

2013年省赛H题你不能每次都快速幂算A^x,优化就是预处理,把10^9预处理成10^5和10^4.想法真的是非常巧妙啊N=100000构造两个数组,f1[N],间隔为Af2[1e4]间隔为A^N,中间用f1来填补f[x]=f1[x%N]*f2[x/N]%P; #include<iostream> #include<cstdio> #include<queue> #include<algorithm> #include<cmath> #inclu…

2013年省赛I题判断单向联通,用bfs剪枝:从小到大跑,如果遇到之前跑过的点(也就是编号小于当前点的点),就o(n)传递关系. bfs #include<iostream> #include<cstdio> #include<queue> #include<algorithm> #include<cmath> #include<ctime> #include<set> #include<map> #inclu…

2018ACM-ICPC南京现场赛D题-Country Meow Problem D. Country Meow Input file: standard input Output file: standard output In the 24th century, there is a country somewhere in the universe, namely Country Meow. Due to advanced technology, people can easily tra…

Cut the Cake Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 300    Accepted Submission(s): 135 Problem Description MMM got a big big big cake, and invited all her M friends to eat the cake toge…

Flyer Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 386    Accepted Submission(s): 127 Problem Description The new semester begins! Different kinds of student societies are all trying to adver…

Walk Through Squares Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 200    Accepted Submission(s): 57 Problem Description   On the beaming day of 60th anniversary of NJUST, as a military colleg…

Poker Shuffle Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 95    Accepted Submission(s): 24 Problem Description Jason is not only an ACMer, but also a poker nerd. He is able to do a perfect s…

Divide Groups Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 423    Accepted Submission(s): 161 Problem Description   This year is the 60th anniversary of NJUST, and to make the celebration mor…

Count The Pairs Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 277    Accepted Submission(s): 150 Problem Description   With the 60th anniversary celebration of Nanjing University of Science…