Conturbatio Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 786    Accepted Submission(s): 358 Problem Description There are many rook on a chessboard, a rook can attack the row and column it be…

摘自题解: 若没有边权,则对点权从大到小排序即可.. 考虑边,将边权拆成两半加到它所关联的两个点的点权中即可. ..因为当两个人分别选择不同的点时,这一权值将互相抵消. 代码如下: #include <cstdio> #include <cstring> #include <algorithm> #define LL long long #define INF 0x7fffffff #define M 100010 using namespace std; int co…

假设删除第k位,把整数A表示成如下形式: A = a * 10^(k+1) + b * 10 ^k + c; 则: B = a * 10^k + c; N = A + B = (11*a+b)*10^k + 2*c; 显然: 11*a+b = N / (10^k) 2*c = N % (10^k) 但是c有可能产生进位,产生的影响为: 11*a+b+1 = N/(10^k)[b+1最多为10,不会影响到11*a的值] 2*c = N % (10^k) + 10^k; 把这两种情况分别考虑一下.…

HDU 4041 Eliminate Witches! (模拟题 ACM ICPC 2011 亚洲北京赛区网络赛题目) Eliminate Witches! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 863    Accepted Submission(s): 342 Problem Description Kaname Mado…

题目链接:传送门 题目: E. Vasya and a Tree time limit per test seconds memory limit per test megabytes input standard input output standard output Vasya has a tree consisting of n vertices with root . At first all vertices has written on it. Let d(i,j) be the…

传送门 解题思路 比较好想的思路题.首先肯定要把原序列转化一下,大于\(k\)的变成\(1\),小于\(k\)的变成\(-1\),然后求一个前缀和,还要用\(cnt[]\)记录一下前缀和每个数出现了几次,然后统计答案的时候从\(1\)循环到\(k\),每次转化为前缀和相减即可. 代码 #include<iostream> #include<cstdio> #include<cstring> using namespace std; const int MAXN = 10…

题目链接 F(N) Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4579 Accepted Submission(s): 1610 Problem Description Giving the N, can you tell me the answer of F(N)? Input Each test case contains a si…

久しぶり! 发现的一道有意思的题,想了半天都没有找到规律,结果竟然是思路题..(在大佬题解的帮助下) 原题戳>>https://www.51nod.com/onlineJudge/questionCode.html#!problemId=1305<< 有这样一段程序,fun会对整数数组A进行求值,其中Floor表示向下取整: fun(A)     sum = 0     for i = 1 to A.length         for j = i+1 to A.length  …

思路: 思路题 题目诡异地给了一组可行匹配 肯定有用啊-. 就把那组可行的解 女向男连一条有向边 如果男喜欢女 男向女连一条有向边 跑一边Tarjan就行了 (这个时候 环里的都能选 "增广环"嘛) 嗯 就搞定了 //By SiriusRen #include <cstdio> #include <cstring> #include <algorithm> using namespace std; #define N 4005 int first[N…

传送门 解题思路 比较好想的一道思路题,结果有个地方没开\(long\) \(long\) \(wa\)了三次..其实就是模仿一下树链剖分,重新定义重儿子,一个点的重儿子为所有儿子中到叶节点权值最大的点,然后就和树链剖分一样\(dfs\)一遍,把那些链的顶端的\(sum\)值放到一个数组排个序. 代码 #include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include&…

BestCoder Sequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1614    Accepted Submission(s): 566 Problem Description Mr Potato is a coder.Mr Potato is the BestCoder. One night, an amazing…

分析:就是判断简单的前缀有没有相同,注意下自身是m的倍数,以及vis[0]=true; #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <iostream> #include <algorithm> #include <map> #include <queue> #include <vect…

题意:如果一个点,则这点的横竖皆被占领,询问矩阵是否全被占领. 思路:将被占领的x,y标记为1,用x表示1 - i的和 如果x轴的差为 x2 - x1 + 1则表示全被占领,y轴同理 #include <cstdio> #include <cstring> #include <algorithm> #include <functional> #include <vector> #include <queue> typedef long…

Problem Description 某国为了防御敌国的导弹袭击,发展出一种导弹拦截系统.但是这种导弹拦截系统有一个缺陷:虽然它的第一发炮弹能够到达任意的高度,但是以后每一发炮弹都不能超过前一发的高度.某天,雷达捕捉到敌国的导弹来袭.由于该系统还在试用阶段,所以只有一套系统,因此有可能不能拦截所有的导弹.怎么办呢?多搞几套系统呗!你说说倒蛮容易,成本呢?成本是个大问题啊.所以俺就到这里来求救了,请帮助计算一下最少需要多少套拦截系统.   Input 输入若干组数据.每组数据包括:导弹总个数(正…

中位数计数 Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 909    Accepted Submission(s): 346 Problem Description 中位数定义为所有值从小到大排序后排在正中间的那个数,如果值有偶数个,通常取最中间的两个数值的平均数作为中位数. 现在有n个数,每个数都是独一无二的,求出每个数在多少个包…

Recently, iSea went to an ancient country. For such a long time, it was the most wealthy and powerful kingdom in the world. As a result, the people in this country are still very proud even if their nation hasn't been so wealthy any more. The merchan…

Building Blocks Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 2209    Accepted Submission(s): 509 Problem Description After enjoying the movie,LeLe went home alone. LeLe decided to build block…

参考:http://www.cnblogs.com/oyking/p/3323306.html 相当不错的思路,膜拜之~ 个人理解改日补充. #include <cstdio> #include <cstring> #include <cstdlib> #include <algorithm> #define lson l, m, rt << 1 #define rson m + 1, r, rt << 1 | 1 #define l…

Operation the Sequence Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 842    Accepted Submission(s): 288 Problem Description You have an array consisting of n integers: a1=1,a2=2,a3=3,…,an=n. T…

Clarke and puzzle Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 673    Accepted Submission(s): 223 Problem Description Clarke is a patient with multiple personality disorder. One day, Clarke s…

Boring count Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 932    Accepted Submission(s): 382 Problem Description You are given a string S consisting of lowercase letters, and your task is cou…

中位数计数 Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 999    Accepted Submission(s): 383 Problem Description 中位数定义为所有值从小到大排序后排在正中间的那个数,如果值有偶数个,通常取最中间的两个数值的平均数作为中位数. 现在有n个数,每个数都是独一无二的,求出每个数在多少个包…

题目: Given a sequence 1,2,3,......N, your job is to calculate all the possible sub-sequences that the sum of the sub-sequence is M. 思路: 刚开始做写了一个尺取法,脑抽的一批.(这几天心情真的是颓的很,连带脑瓜也不好用了,抓紧调整~~~~) 其实可以根据等差数列求和公式求出这个子序列的长度的一个大致的范围为k=sqrt(2*m),然后根据m和k求出首项a1,然后将a1…

题意:在[a,b]  [c,d] 之间,和模p等于m的对数 详见代码 #include <stdio.h> #include <algorithm> #include <string.h> #include<cmath> #define LL long long using namespace std; int T; LL a,b,c,d,p,m; LL gcd(LL a, LL b) { return b ? gcd(b, a % b) : a; } LL…

考虑对立情况,不漂亮的串的形式必然为GGGGR……R……RGGGG 相邻R之间的距离为奇数且相等. #include <cstdio> #include <cstring> #include <cstdlib> #define LL long long int using namespace std; ; ; char str[MAXN]; int main() { int T; scanf( "%d", &T ); while ( T--…

Select Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1599    Accepted Submission(s): 443 Problem Description One day, Dudu, the most clever boy, heard of ACM/ICPC, which is a very interesting…

Goffi and Squary Partition Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1171    Accepted Submission(s): 402 Problem Description Recently, Goffi is interested in squary partition of integers.…

Poor Hanamichi Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 893    Accepted Submission(s): 407 Problem Description Hanamichi is taking part in a programming contest, and he is assigned to sol…

Arithmetic Sequence Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1445    Accepted Submission(s): 632 Problem Description A sequence b1,b2,⋯,bn are called (d1,d2)-arithmetic sequence if and on…

Conturbatio Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 232    Accepted Submission(s): 108 Problem Description There are many rook on a chessboard, a rook can attack the row and column it b…