<题目链接> 题目大意: 给定数字n,让你将其分成合数相加的形式,问你最多能够将其分成几个合数相加. 解题分析: 因为要将其分成合数相加的个数最多,所以自然是尽可能地将其分成尽可能小的合数相加的形式.通过找规律,我们能够发现,所有的偶数都能够分成4和6这两个合数的组合,而所有的奇数,在减去9这个最小的奇合数后,就会变成偶数,然后就是和普通偶数一样的处理方式. 普通偶数的处理方式就是,看他能够分成几个4,如果该偶数不为4的倍数,那么就是将其中的一个4换成6.总的最大合数个数为:$n/4$ 而奇数…

一.题目 Given an integer $n$, Chiaki would like to find three positive integers $x$, $y$ and $z$ such that: $n=x+y+z$, $x\mid n$, $y \mid n$, $z \mid n$ and $xyz$ is maximum.  InputThere are multiple test cases. The first line of input contains an integ…

题目 https://nanti.jisuanke.com/t/17118 题意 有n个点0,1,2...n-1,对于一个点对(i,j)满足i<j,那么连一条边,边权为i xor j,求0到n-1的最大流,结果取模,n<=1e18 分析 可以写个最大流对数据找规律,但没找出来…… 然后只能取分析了,首先最大流等价于最小割 明确一定,0->n-1这个要先割掉 然后我们贪心,希望有一些点割掉与0相连的边,一些点割掉与n-1相连的边 我们去观察每个点与0相连和与n-1相连的两条边权值,容易发现…

Educational Codeforces Round 78 (Rated for Div. 2) 1278B - 6 B. A and B  time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You are given two integers aa and bb. You can perform a sequence of…

A. Little Pony and Expected Maximum Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/453/problem/A Description Twilight Sparkle was playing Ludo with her friends Rainbow Dash, Apple Jack and Flutter Shy. But she kept losing.…

题意:给出一个字符串,有两种操作: 1.插入一个数字  2.交换两个字符   问最少多少步可以把该字符串变为一个后缀表达式(操作符只有*). 解法:仔细观察,发现如果数字够的话根本不用插入,数字够的最低标准为'*'的个数+1,因为最优是 '12*3*..' 这种形式,所以先判断够不够,不够就补,然后从左往右扫一遍,如果某个时刻Star+1>Num,那么从开始到这一段是不合法的,要把那个'*'与后面的一个数字交换,此时Star--,Num++.然后步数++.这样得出的结果就是最后的最小步数. 脑子…

Ball Painting 题目连接: http://codeforces.com/gym/100015/attachments Description There are 2N white balls on a table in two rows, making a nice 2-by-N rectangle. Jon has a big paint bucket full of black paint. (Don't ask why.) He wants to paint all the b…

B. Lunch Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100637/problem/B Description The swamp looks like a narrow lane with length n covered by floating leaves sized 1, numbered from 1 to n with a fly sitting on the top of ea…

题目链接:http://codeforces.com/problemset/problem/603/A 题意: 给定一个 $01$ 串,我们“交替子序列”为这个串的一个不连续子序列,它满足任意的两个相邻的数字不相等. 现在,我们要对这个 $01$ 串的某一段非空连续子串进行反转操作,即将这一段上的所有 $0$ 变为 $1$,所有 $1$ 变为 $0$. 然后,求问进行了有且仅有一次的反转操作后,求该串的最长交替子序列的长度. 题解: 首先,对于一个 $01$ 串,对其进行压缩操作,即将所有连续的…

题目链接:http://codeforces.com/contest/872/problem/C 题意: 给你一个数n,问你最多能将n分解成多少个合数之和.(若不能分解,输出-1) 题解: 若要让合数个数最多,则n必定只由4,6,9组成. n由n/4和n%4两部分组成. 四种情况: (1)n%4 == 0: 全分成4就好了,所以ans = n/4 (2)n%4 == 1: 剩下的1要和两个4组合成一个9. 所以如果n/4 >= 2,ans = n/4 - 1 否则ans = -1 (3)n%4…

D. Flowers time limit per test:1.5 seconds memory limit per test:256 megabytes We saw the little game Marmot made for Mole's lunch. Now it's Marmot's dinner time and, as we all know, Marmot eats flowers. At every dinner he eats some red and white flo…

题目传送门 /* 找规律/贪心:ans = n - 01匹配的总数,水 */ #include <cstdio> #include <iostream> #include <algorithm> #include <cstring> #include <cmath> using namespace std; ; const int INF = 0x3f3f3f3f; char s[MAXN]; int main(void) //Codeforce…

C. Maximum splitting time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output You are given several queries. In the i-th query you are given a single positive integer ni. You are to represent ni as…

题目链接: http://codeforces.com/problemset/problem/353/D?mobile=true H. Queue time limit per test 1 secondmemory limit per test 256 megabytes 问题描述 There are n schoolchildren, boys and girls, lined up in the school canteen in front of the bun stall. The b…

A. Hanoi tower Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100114 Description you the conditions of this task. There are 3 pivots: A, B, C. Initially, n disks of different diameter are placed on the pivot A: the smallest dis…

题目链接: https://codeforces.com/contest/166/problem/E 题目: 题意: 给你一个三菱锥,初始时你在D点,然后你每次可以往相邻的顶点移动,问你第n步回到D点的方案数. 思路: 打表找规律得到的序列是0,3,6,21,60,183,546,1641,4920,14763,通过肉眼看或者oeis可以得到规律为. dp计数:dp[i][j]表示在第i步时站在位置j的方案数,j的取值为[0,3],分别表示D,A,B,C点,转移方程肯定是从其他三个点转移. 代码…

题目传送门 /* 找规律,水 */ #include <cstdio> #include <iostream> #include <algorithm> #include <cstring> #include <cmath> using namespace std; ; const int INF = 0x3f3f3f3f; ]; int main(void) //Codeforces Round #309 (Div. 2) A. Kyoya a…

题目描述: Little Elephant and Interval time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output The Little Elephant very much loves sums on intervals. This time he has a pair of integers l and r (l ≤ r…

Codeforces 题目传送门 & 洛谷题目传送门 蠢蠢的我竟然第一眼想套通项公式?然鹅显然 \(5\) 在 \(\bmod 10^{13}\) 意义下并没有二次剩余--我真是活回去了... 考虑打表找规律(u1s1 这是一个非常有用的技巧,因为这个 \(10^{13}\) 给的就很灵性,用到类似的技巧的题目还有这个,通过对这些模数的循环节打表找出它们的共同性质,所以以后看到什么特殊的数据或者数据范围特别大但读入量 \(\mathcal O(1)\) 的题(比如 CF838D)可以考虑小数据打…

题目传送门 /* 水题 找规律输出 */ #include <cstdio> #include <iostream> #include <cstring> #include <string> #include <map> #include <algorithm> #include <vector> #include <set> #include <cmath> using namespace std…

BOPCTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=86686#problem/D Description You invented a new chess figure and called it BOPC. Suppose it stands at the square grid at the point with coordinates …

题目:Click here 题意:给定数列满足求f(n)mod(1e9+7). 分析:规律题,找规律,特别注意负数取mod. #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> using namespace std; ; ; int x, y, n; ]; int main() { while( ~scanf…

题意:给出一个n,生成n的所有全排列,将他们按顺序前后拼接在一起组成一个新的序列,问有多少个长度为n的连续的子序列和为(n+1)*n/2 题解:由于只有一个输入,第一感觉就是打表找规律,虽然表打出来了,但是依然没有找到规律...最后看了别人的题解才发现 ans [ 3 ] = 1*2*3 + ( ans [ 2 ] - 1 ) * 3 ans[ 4 ] = 1*2*3*4 + ( ans[ 3 ] - 1 ) * 4 感觉每次离成功就差一点点!!! #include<bits/stdc++.h>…

地址: 题目: C. Maximum splitting time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output You are given several queries. In the i-th query you are given a single positive integer ni. You are to represe…

Description You are given an integer N. Consider all possible segments (线段,划分)on the coordinate axis with endpoints at integer points with coordinates between 0 and N, inclusive; there will be  of them. You want to draw these segments in several laye…

D. Roman Digits time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Let's introduce a number system which is based on a roman digits. There are digits I, V, X, L which correspond to the numbers…

打表找规律即可. 1,1,2,2,2,3,3,3,3,4,4,4,4,4... 注意打表的时候,sg值不只与剩下的石子数有关,也和之前取走的方案有关. //#include<cstdio> //#include<set> //#include<cstring> //using namespace std; //bool vis[16]; //int n,SG[16][1<<16]; //int sg(int x,int moved) //{ // if(SG…

一看就是找规律的题.只要熟悉异或的性质,可以秒杀. 为了防止忘记异或的规则,可以把异或理解为半加运算:其运算法则相当于不带进位的二进制加法. 一些性质如下: 交换律: 结合律: 恒等律: 归零律: 典型应用:交换a和b的值:a=a^b^(b=a); #include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<cmath> #include<…

刚开始以为是搜索白忙活了原来是个简单的找规律,以后要多想啊 此题是两马同时跳 A. Two Semiknights Meet time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output A boy Petya loves chess very much. He even came up with a chess piece of his o…

题目: 解题思路 这题就是0,1,2...n-1总共n个数字形成的最小生成树. 我们可以发现,一个数字k与比它小的数字形成的异或值,一定可以取到k与所有正整数形成的异或值的最小值. 要计算n个数字的情况我们可以通过n-1个数字的情况得来,意为前n-1个数字的最小生成树已经生成好了,我们需要给第n个数字连一条边,使新的树为n个数字的最小生成树. 通过找规律我们可以发现: 每隔2个数字多一个权值为1的边. 每隔4个数字多一个权值为2的边. 每隔8个数字多一个权值为4的边. …… 每隔2^n个数字多一…