题目链接:http://codeforces.com/problemset/problem/450/B 题意很好懂,矩阵快速幂模版题. /* | 1, -1 | | fn | | 1, 0 | | fn-1 | */ #include <iostream> #include <cstdio> #include <cstring> using namespace std; typedef __int64 LL; LL mod = 1e9 + ; struct data {…

题目链接:http://codeforces.com/problemset/problem/450/B B. Jzzhu and Sequences time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Jzzhu has invented a kind of sequences, they meet the following pr…

Jzzhu has invented a kind of sequences, they meet the following property: You are given x and y, please calculate fn modulo 1000000007 (109 + 7). Input The first line contains two integers x and y (|x|, |y| ≤ 109). The second line contains a single i…

B解题报告 算是规律题吧,,,x y z -x -y -z 注意的是假设数是小于0,要先对负数求模再加模再求模,不能直接加mod,可能还是负数 给我的戳代码跪了,,. #include <iostream> #include <cstring> #include <cstdio> using namespace std; long long x,y,z; long long n; int main() { cin>>x>>y; cin>&g…

主题链接:http://codeforces.com/problemset/problem/449/A ---------------------------------------------------------------------------------------------------------------------------------------------------------- 欢迎光临天资小屋:http://user.qzone.qq.com/593830943…

题目链接:http://codeforces.com/problemset/problem/450/A ---------------------------------------------------------------------------------------------------------------------------------------------------------- 欢迎光临天资小屋:http://user.qzone.qq.com/593830943…

题目链接:http://codeforces.com/problemset/problem/449/C 给你n个数,从1到n.然后从这些数中挑选出不互质的数对最多有多少对. 先是素数筛,显然2的倍数的个数是最多的,所以最后处理.然后处理3,5,7,11...的倍数的数,之前已经挑过的就不能再选了.要是一个素数p的倍数个数是奇数,就把2*p给2 的倍数.这样可以满足p倍数搭配的对数是最优的.最后处理2的倍数就行了. #include <bits/stdc++.h> using namespace…

D - Jzzhu and Numbers 这个容斥没想出来... 我好菜啊.. f[ S ] 表示若干个数 & 的值 & S == S得 方案数, 然后用这个去容斥. 求f[ S ] 需要用SOSdp #include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define PLL pair<LL, LL> #define…

C. Jzzhu and Chocolate time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Jzzhu has a big rectangular chocolate bar that consists of n × m unit squares. He wants to cut this bar exactly k time…

A. Jzzhu and Children time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output There are n children in Jzzhu's school. Jzzhu is going to give some candies to them. Let's number all the children from…

题意:n个城市,中间有m条道路(双向),再给出k条铁路,铁路直接从点1到点v,现在要拆掉一些铁路,在保证不影响每个点的最短距离(距离1)不变的情况下,问最多能删除多少条铁路 分析:先求一次最短路,铁路的权值大于该点最短距离的显然可以删去,否则将该条边加入图中,再求最短路,记录每个点的前一个点,然后又枚举铁路,已经删去的就不用处理了,如果铁路权值大于该点最短距离又可以删去,权值相等时,该点的前一个点如果不为1,则这个点可以由其他路到达,这条铁路又可以删去. 由于本题中边比较多,最多可以有8x10^…

E. Sasha and Array time limit per test 5 seconds memory limit per test 256 megabytes input standard input output standard output Sasha has an array of integers a1, a2, ..., an. You have to perform m queries. There might be queries of two types: 1 l r…

今天老师(orz sansirowaltz)让我们做了很久之前的一场Codeforces Round #257 (Div. 1),这里给出A~C的题解,对应DIV2的C~E. A.Jzzhu and Chocolate time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Jzzhu has a big rectangular cho…

题目传送门 /* DP:先用l,r数组记录前缀后缀上升长度,最大值会在三种情况中产生: 1. a[i-1] + 1 < a[i+1],可以改a[i],那么值为l[i-1] + r[i+1] + 1 2. l[i-1] + 1 3. r[i+1] + 1 //修改a[i] */ #include <cstdio> #include <algorithm> #include <cstring> using namespace std; ; const int INF…

B. Jzzhu and Sequences time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Jzzhu has invented a kind of sequences, they meet the following property: You are given x and y, please calculate fn m…

A - Jzzhu and Children 找到最大的ceil(ai/m)即可 #include <iostream> #include <cmath> using namespace std; int main(){ int n,m; cin >> n >> m; ; ; ; i < n; ++ i){ cin >> a; if(maxv <= ceil(a/m)){ maxv = ceil(a/m); maxIdx = i+;…

B. Jzzhu and Sequences time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Jzzhu has invented a kind of sequences, they meet the following property: You are given x and y, please calculate fn m…

Problem A A. Jzzhu and Children time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output There are n children in Jzzhu's school. Jzzhu is going to give some candies to them. Let's number all the chi…

Jzzhu has invented a kind of sequences, they meet the following property: You are given x and y, please calculate fn modulo 1000000007 (109 + 7). Input The first line contains two integers x and y (|x|, |y| ≤ 109). The second line contains a single i…

题目链接: B. Jzzhu and Sequences time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Jzzhu has invented a kind of sequences, they meet the following property: You are given x and y, please calculat…

D. Jzzhu and Cities time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Jzzhu is the president of country A. There are n cities numbered from 1 to n in his country. City 1 is the capital of A.…

E. Jzzhu and Apples time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Jzzhu has picked n apples from his big apple tree. All the apples are numbered from 1 to n. Now he wants to sell them to…

A. Jzzhu and Children time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output There are n children in Jzzhu's school. Jzzhu is going to give some candies to them. Let's number all the children from…

解题报告 没什么好说的,大于m的往后面放,,,re了一次,,, #include <iostream> #include <cstdio> #include <cstring> #include <cmath> using namespace std; struct node { int x,cd; }num[1000000]; int main() { int n,m,c; cin>>n>>m; int i; for(i=0;i&l…

思路:定义f(x)为 Ai & x==x  的个数,g(x)为x表示为二进制时1的个数,最后答案为    .为什么会等于这个呢:运用容斥的思想,如果 我们假设 ai&x==x 有f(x)个,那么 这f(x)个 组成集合的子集 & 出来是 >=x那么我们要扣掉>x的 ...  因为这里我们要求的是 & 之后等于0 一开始1个数为0那么就是 1个数为偶数时加上去,  为奇数时减掉了. 那么就剩下求f(x)    .我们把A[i]和x的二进制 分成  前 (20-k)…

题意:给力一张无向图,有一些边是正常道路,有一些边是铁路,问最多能删除几条铁路使得所有点到首都(编号为1)的最短路长度不变. 思路:求不能删除的铁路数,总数减掉就是答案.先求出首都到所有点的最短路,求完最短路后,枚举除首都外所有点,如果这个点被更新的边中只有铁路,那么就有一条铁路不能删除. 注意:这里求最短路一开始用SPFA在第45个点TLE,最后换成带堆优化Dijkstra #include<cstring> #include<algorithm> #include<cst…

题目链接: http://www.codeforces.com/contest/446/problem/A 题解: dp1[x]表示以x结尾的最大严格升序连续串,dp2[x]表示以x开头的最大严格升序连续串 #include<iostream> #include<cstdio> #include<cstring> #include<map> #include<list> #include<stack> #include<algo…

A. DZY Loves Sequences 题目连接: http://www.codeforces.com/contest/446/problem/A Description DZY has a sequence a, consisting of n integers. We'll call a sequence ai, ai + 1, ..., aj (1 ≤ i ≤ j ≤ n) a subsegment of the sequence a. The value (j - i + 1) d…

https://codeforces.com/contest/1330/problem/D 题目大意:给出一个限制 d 与模数 m ,求出可以构造出的满足条件的数组 a 的个数,需要满足以下条件:    1.数组 a 的长度大于等于 1     2.数组 a 严格递增    3.任意的ai <=d且>=1    4.对于数组 a ,需要构造出一个数组 b :满足当 i == 1 时:b[ 1 ] = a[ 1 ], i > 1 时:b[ i ] = b[ i - 1 ] XOR a[ i…

解题报告:输入一个数列,选取一个子数列,要求最多只能改动这个子数列中的一个数,使得这个子数列是严格的升序的(严格升序没有相等的) 我的做法是,第一步把这个 数列的每个升序的子数列都找出来,然后看这些子数列能不能和跟它相邻的升序的子数列连接起来. #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<cmath> using namesp…