记录一下每个对角线上有几个,然后就可以算了 #include<cstdio> #include<cstring> #include<cmath> #include<ctime> #include<vector> #include<algorithm> using namespace std; +; int n; long long w1[maxn]; long long w2[maxn]; long long ans; int mai…

B. Wet Shark and Bishops time limit per test: 2 seconds memory limit per test: 256 megabytes input: standard input output: standard output Today, Wet Shark is given n bishops on a 1000 by 1000 grid. Both rows and columns of the grid are numbered from…

对角线 x1+y1=x2+y2 或者x1-y1=x2-y2 #include<iostream> #include<string> #include<algorithm> #include<cstdlib> #include<cstdio> #include<set> #include<map> #include<vector> #include<cstring> #include<stack…

B. Wet Shark and Bishops time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Today, Wet Shark is given n bishops on a 1000 by 1000 grid. Both rows and columns of the grid are numbered from 1 t…

Today, Wet Shark is given n bishops on a 1000 by 1000 grid. Both rows and columns of the grid are numbered from 1 to 1000. Rows are numbered from top to bottom, while columns are numbered from left to right. Wet Shark thinks that two bishops attack e…

http://codeforces.com/problemset/problem/621/E E. Wet Shark and Blocks time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output There are b blocks of digits. Each one consisting of the same n digit…

E. Wet Shark and Blocks time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output There are b blocks of digits. Each one consisting of the same n digits, which are given to you in the input. Wet Sha…

题 题意 1000*1000的格子里,给你n≤200 000个点的坐标,求有多少对在一个对角线上. 分析 如果求每个点有几个共对角线的点,会超时. 考虑到对角线总共就主对角线1999条+副对角线1999条,我们可以求每个对角线有几对点. 同一条主对角线上的元素有a[i]个,就有C(a[i],2)对点: 同一条副对角线上的元素有b[i]个,就有C(b[i],2)对点. 读入x和y后, x+y相同的就在同一副对角线,x+y范围是(2,2000), x-y相同的就是同一主对角线,x-y范围是(-999…

题意:处在同一对角线上的主教(是这么翻译没错吧= =)会相互攻击 求互相攻击对数 由于有正负对角线 因此用两个数组分别保存每个主教写的 x-y 和 x+y 然后每个数组中扫描重复数字k ans加上kC2就行了 wa了两发的原因是没考虑到如果整个数组都是重复的 那要最后额外加一次 #include <cstdio> #include <cmath> #include <cstring> #include <queue> #include <vector&…

B. Chocolate time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Bob loves everything sweet. His favorite chocolate bar consists of pieces, each piece may contain a nut. Bob wants to break the…

方法可以转化一下,先计算每一个鲨鱼在自己范围内的数能被所给素数整除的个数有几个,从而得到能被整除的概率,设为f1,不能被整除的概率设为f2. 然后计算每相邻两只鲨鱼能获得钱的期望概率,f=w[id1].f1*w[id2].f2+w[id1].f2*w[id2].f1+w[id1].f1*w[id2].f1; f*2000就是这两只鲨鱼能获得的期望金钱,然后枚举一下所有相邻的鲨鱼,累加即可. #include<cstdio> #include<cstring> #include<…

水题 #include<cstdio> #include<cstring> #include<cmath> #include<ctime> #include<vector> #include<algorithm> using namespace std; int n; +]; int main() { scanf("%d",&n); ;i<n;i++) scanf("%lld",&…

题意: 有b个blocks,每个blocks都有n个相同的0~9的数字,如果从第一个block选1,从第二个block选2,那么就构成12,问对于给定的n,b有多少种构成方案使最后模x的余数为k. 分析: dp+矩阵快速幂. 假如现在的数是m,模x余数是n,那么再从下一个block中选一个数a,a模x余数为b,那么新的数的余数就为(m∗10+a)%x,也就是(n∗10+b)%x,所以实际上我们只需要直接对余数进行操作.容易得到状态转移方程,其中dp[i][j]表示从第i个block中选择一个数后…

 B. Wet Shark and Bishops time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Today, Wet Shark is given n bishops on a 1000 by 1000 grid. Both rows and columns of the grid are numbered from…

题目链接: http://codeforces.com/problemset/problem/621/E E. Wet Shark and Blocks time limit per test2 secondsmemory limit per test256 megabytes 问题描述 There are b blocks of digits. Each one consisting of the same n digits, which are given to you in the inp…

题 Today, Wet Shark is given n integers. Using any of these integers no more than once, Wet Shark wants to get maximum possible even (divisible by 2) sum. Please, calculate this value for Wet Shark. Note, that if Wet Shark uses no integers from the n …

题 There are n sharks who grow flowers for Wet Shark. They are all sitting around the table, such that sharks i andi + 1 are neighbours for all i from 1 to n - 1. Sharks n and 1 are neighbours too. Each shark will grow some number of flowers si. For i…

time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output There are b blocks of digits. Each one consisting of the same n digits, which are given to you in the input. Wet Shark must choose exactly one d…

C. Wet Shark and Flowers time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output There are n sharks who grow flowers for Wet Shark. They are all sitting around the table, such that sharks i and i …

A. Wet Shark and Odd and Even time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Today, Wet Shark is given n integers. Using any of these integers no more than once, Wet Shark wants to get ma…

题意: 不概括了..太长了.. 额第一次做这种问题 算是概率dp吗? 保存前缀项中第一个和最后一个的概率 然后每添加新的一项 就解除前缀和第一项和最后一项的关系 并添加新的一项和保存的两项的关系 这里关系指的是两者相邻会产生的额外收入(其中一个满足条件就能得到 因此公式是 2000 * (rate[a] * rate[b] + rate[a] * ( 1 - rate[b]) + rate[b] * (1 - rate[a])) 至于一开始为什么老是调不过去呢..我发现添加第三项的时候前两项是不…

题意是得到最大的偶数和 解决办法很简单 排个序 取和 如果是奇数就减去最小的奇数 #include <cstdio> #include <cmath> #include <cstring> #include <queue> #include <vector> #include <algorithm> #define INF 0x3f3f3f3f #define mem(str,x) memset(str,(x),sizeof(str)…

题目大意:有m (m<=1e9) 个相同的块,每个块里边有n个数,每个数的范围是1-9,从每个块里边取出来一个数组成一个数,让你求组成的方案中 被x取模后,值为k的方案数.(1<=k<x<=100) 思路:刚开始把m看成了1e5,写了一发数位dp,果断RE,然后我在考虑从当块前状态转移到下一个块状态模数的改变都用的是一套规则,因为 每个块里面的数字都是一样的,那么我们就可以很开心地构造出一个x*x的矩阵进行快速幂啦. #include<bits/stdc++.h> #d…

http://codeforces.com/problemset/problem/735/C 题意:有n个人打锦标赛,淘汰赛制度,即一个人和另一个人打,输的一方出局.问这n个人里面冠军最多能赢多少场,其中一个人和另一个人能打比赛当且仅当这两个人赢的局数相差不超过1. 思路:比赛的时候不会做..直接log2(n)交,果断错了.看题解:因为限制条件两个人能比赛当且仅当他们赢得局数相差不超过1,设F[x]为冠军赢x盘的时候需要的总人数,那么有这样的递推式:F[x] = F[x-1] + F[x-2].…

D. Spongebob and Squares Spongebob is already tired trying to reason his weird actions and calculations, so he simply asked you to find all pairs of n and m, such that there are exactly x distinct squares in the table consisting of n rows and m colum…

Problem D"Decoding Task" Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100002 Description In the near future any research and publications about cryptography are outlawed throughout the world on the grounds of national se…

http://codeforces.com/contest/716/problem/C 题目大意:感觉这道题还是好懂得吧. 思路:不断的通过列式子的出来了.首先我们定义level=i, uplevel = i + 1,目前的uplevel,然后我们可以知道,之前求出来的restgrade%level = 0,于是我们可以列出一个式子sqrt(k1 * i + grade) = uplevel * k.因为grade也是i的倍数,假定grade = k2 * i,所以我们写成sqrt((k1 +…

D. Winter is here time limit per test:3 seconds memory limit per test:256 megabytes input:standard input output:standard output Winter is here at the North and the White Walkers are close. John Snow has an army consisting of n soldiers. While the res…

A continued fraction of height n is a fraction of form . You are given two rational numbers, one is represented as  and the other one is represented as a finite fraction of height n. Check if they are equal. Input The first line contains two space-se…

https://vjudge.net/problem/48715/origin 题意:给出必定含1689四个数字的字符串,随意交换位置构造出能被7整除的数. 分析:数学思维题.观察发现1689的排列与7的余数恰好是0...6,那么利用这个性质去与串中其他数字相补出能被7整除的数,把1689固定在后四位,把0独自提取出来,其他数字为一组,把它们的数值*10000后mod7,根据得到的余数选择相应的1689排列,最后把0放在末尾(对结果正确性没有影响).代码: #include <cstdio>…