Rikka with Chess Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 177    Accepted Submission(s): 161 Problem Description Yuta gives Rikka a chess board of size n×m. As we all know, on a chess boa…

KK's Point Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 644    Accepted Submission(s): 220 Problem Description Our lovely KK has a difficult mathematical problem:He points N(2≤N≤105) points o…

KK's Steel Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 350    Accepted Submission(s): 166 Problem Description Our lovely KK has a difficult mathematical problem:he has a N(1≤N≤1018) meters s…

Rikka with Graph  Accepts: 123  Submissions: 525  Time Limit: 2000/1000 MS (Java/Others)  Memory Limit: 65536/65536 K (Java/Others) 问题描述 众所周知,萌萌哒六花不擅长数学,所以勇太给了她一些数学问题做练习,其中有一道是这样的: 给出一张 nn 个点 n+1n+1 条边的无向图,你可以选择一些边(至少一条)删除. 现在勇太想知道有多少种方案使得删除之后图依然联通.…

Rikka with Phi  Accepts: 5  Submissions: 66  Time Limit: 16000/8000 MS (Java/Others)  Memory Limit: 131072/131072 K (Java/Others) Problem Description Rikka and Yuta are interested in Phi function (which is known as Euler's totient function). Yuta giv…

Rikka with Chess  Accepts: 393  Submissions: 548  Time Limit: 2000/1000 MS (Java/Others)  Memory Limit: 65536/65536 K (Java/Others) 问题描述 一个n \times mn×m的黑白相间的棋盘,每次可以选择一个矩形把其中的所有格子反色.问把所有格子变成一种颜色时的最少操作次数. 输入描述 第一行一个整数 T(T \leq 10)T(T≤10) 表示数据组数. 每组数据有…

A - Elections 题意: 每一场城市选举的结果,第一关键字是票数(降序),第二关键字是序号(升序),第一位获得胜利. 最后的选举结果,第一关键字是获胜城市数(降序),第二关键字是序号(升序),第一位获得胜利. 求最后选举获胜者. 思路: 直接模拟就可以. 代码: /* * @author FreeWifi_novicer * language : C++/C */ #include<cstdio> #include<iostream> #include<cstrin…

http://acm.hdu.edu.cn/showproblem.php?pid=5424 哈密顿通路:联通的图,访问每个顶点的路径且只访问一次 n个点n条边 n个顶点有n - 1条边,最后一条边的连接情况: (1)自环(这里不需要考虑): (2)最后一条边将首和尾连接,这样每个点的度都为2: (3)最后一条边将首和除尾之外的点连接或将尾和出尾之外的点连接,这样相应的首或尾的度最小,度为1: (4)最后一条边将首和尾除外的两个点连接,这样就有两个点的度最小,度都为1 如果所给的图是联通的话,那…

Baby Ming and phone number Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1501    Accepted Submission(s): 399 Problem Description Baby Ming collected lots of cell phone numbers, and he wants to…

题意: 给出一张 nnn 个点 n+1n+1n+1 条边的无向图,你可以选择一些边(至少一条)删除. 分析: 一张n个点图,至少n-1条边才能保证联通 所以可以知道每次可以删去1条边或者两条边 一开始看了题解,我搞不出来[每次删去1条或者2条] T^T就是那么弱,妈的 然后判断是否联通,DFS联通图可以,但是好麻烦的 所以想到判断一个集合,用并查集,妥妥的 #include <iostream> #include <stdio.h> #include <stdlib.h>…

1001 Rikka with Chess ans = n / 2 + m / 2 1002 Rikka with Graph 题意:n + 1条边,问减去至少一条使剩下的图连通的方案数. 分析:原来暴力选一条或两条就行了,脑子笨了.判连通用BFS或并查集,此题并查集更好 #include <cstdio> #include <cmath> #include <cstring> #include <algorithm> #include <vector…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5174 题目意思:给出 n 个人坐的缆车值,假设有 k 个缆车,缆车值 A[i] 需要满足:A[i−1]<A[i]<A[i+1](1<i<K).现在要求的是,有多少人满足,(他坐的缆车的值 + 他左边缆车的值) % INT_MAX == 他右边缆车的值. 首先好感谢出题者的样例三,否则真的会坑下不少人.即同一部缆车可以坐多个人.由于缆车的值是唯一的,所以可以通过排序先排出缆车的位置.求出…

点我看题目 A. Mashmokh and Lights time limit per test:1 secondmemory limit per test:256 megabytesinput:standard inputoutput:standard output Mashmokh works in a factory. At the end of each day he must turn off all of the lights. The lights on the factory a…

转载请注明出处: http://www.cnblogs.com/fraud/          ——by fraud Gunner Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Problem Description Long long ago, there is a gunner whose name is Jack. He likes to go hunting very…

2.2.2017 9:35~11:35 A - Taymyr is calling you 直接模拟 #include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> using namespace std; typedef long long ll; ; inline int read(){ ,f=; ; c=getcha…

---恢复内容开始--- Kurt reaches nirvana when he finds the product of all the digits of some positive integer. Greater value of the product makes the nirvana deeper. Help Kurt find the maximum possible product of digits among all integers from 1 to n. Input…

题目:http://codeforces.com/problemset/problem/584/D 思路: 关于偶数的哥德巴赫猜想:任一大于2的偶数都可写成两个素数之和. 关于奇数的哥德巴赫猜想:任一大于7的奇数都可写成三个质数之和的猜想. #include<cstdio> #include<cstring> #include<cmath> #define ll long long using namespace std; bool isprime(ll n) { ;i…

题目 传送门:QWQ A:A - If at first you don't succeed... 分析: 按照题意模拟 代码: #include <bits/stdc++.h> using namespace std; int main(){ int a,b,c,n; scanf("%d%d%d%d",&a,&b,&c,&n); int ans=n-a-b+c; ; ) printf("%d\n",ans); else…

题目 传送门:QWQ A:A - Hit the Lottery 分析: 大水题 模拟 代码: #include <bits/stdc++.h> using namespace std; int main(){ int n; ;scanf("%d",&n); ){ x++; n-=; } ){ x++; n-=; } ){ x++; n-=; } ){ x++; n-=; } printf("%d\n",x+n); } B:B - World C…

Link~ 题面差评,整场都在读题 A 根据奇偶性判断一下即可. #include<bits/stdc++.h> #define ll long long #define N #define rep(i,a,n) for (int i=a;i<=n;i++) #define per(i,a,n) for (int i=n;i>=a;i--) #define inf 0x3f3f3f3f #define pb push_back #define mp make_pair #defin…

Baby Ming and Weight lifting Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 681    Accepted Submission(s): 280 Problem Description Baby Ming is fond of weight lifting. He has a barbell pole(the…

tree Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 863    Accepted Submission(s): 409 Problem Description There is a tree(the tree is a connected graph which contains n points and n−1 edges),t…

Shortest Path  Accepts: 40  Submissions: 610  Time Limit: 4000/2000 MS (Java/Others)  Memory Limit: 131072/131072 K (Java/Others) 问题描述 有一条长度为nn的链. 节点ii和i+1i+1之间有长度为11的边. 现在又新加了3条边, 每条边长度都是1. 给出mm个询问, 每次询问两点之间的最短路. 输入描述 输入包含多组数据. 第一行有一个整数TT, 表示测试数据的组数…

BestCoder Round #64 (div.2) Array 问题描述 Vicky是个热爱数学的魔法师,拥有复制创造的能力. 一开始他拥有一个数列{1}.每过一天,他将他当天的数列复制一遍,放在数列尾,并在两个数列间用0隔开.Vicky想做些改变,于是他将当天新产生的所有数字(包括0)全加1.Vicky现在想考考你,经过100天后,这个数列的前M项和是多少?. 输入描述 输入有多组数据. 第一行包含一个整数T,表示数据组数.T. \left( 1 \leq T \leq 2 * {10}^…

题目链接: huangjing hdu5054 Alice and Bob 思路: 就是(x,y)在两个參考系中的表示演全然一样.那么仅仅可能在这个矩形的中点.. 题目: Alice and Bob Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 216    Accepted Submission(s): 166 Problem De…

Codeforces Beta Round #73 (Div. 2 Only) http://codeforces.com/contest/88 A 模拟 #include<bits/stdc++.h> using namespace std; #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 #define sqr(x) ((x)*(x)) #define pb push_back #define eb emp…

HDOJ5054 Alice and Bob Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 302    Accepted Submission(s): 229 Problem Description Bob and Alice got separated in the Square, they agreed that if they…

LCP Array  Accepts: 131  Submissions: 1352  Time Limit: 4000/2000 MS (Java/Others)  Memory Limit: 131072/131072 K (Java/Others) 问题描述 Peter有一个字符串s=s_{1}s_{2}...s_{n}s=s​1​​s​2​​...s​n​​, 令\text{suff}_i =s_{i}s_{i+1}...s_{n}suff​i​​=s​i​​s​i+1​​...s​n​…

题目链接:BestCoder Round #81 (div.2) 1003 String 题意 中文题,上有链接.就不贴了. 思路 枚举起点i,计算能够达到k个不同字母的最小下标j,则此时有子串len-j个. 将全部起点的值加起来即是结果. 代码 #include<iostream> #include<algorithm> #include<vector> using namespace std; #define LL long long const int MOD =…

题目传送门 /* 题意: 求(n-1)! mod n 数论:没啥意思,打个表能发现规律,但坑点是4时要特判! */ /************************************************ * Author :Running_Time * Created Time :2015-8-15 19:06:12 * File Name :A.cpp ************************************************/ #include <cstdi…