http://codeforces.com/contest/368/problem/D #include <cstdio> #include <cstring> #include <map> #include <vector> #include <algorithm> #define maxn 300000 #define LL long long using namespace std; int n,m,p; LL a[maxn],b[maxn…

B. Sereja ans Anagrams Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/problemset/problem/367/B Description Sereja has two sequences a and b and number p. Sequence a consists of n integers a1, a2, ..., an. Similarly, sequence b con…

Sereja has two sequences a and b and number p. Sequence a consists of n integers a1, a2, ..., an. Similarly, sequence b consists of m integers b1, b2, ..., bm. As usual, Sereja studies the sequences he has. Today he wants to find the number of positi…

Codeforces Round #215 (Div. 1) B:http://codeforces.com/problemset/problem/367/B 题意:给你两个序列a,b,然后给你一个数p,然后让你在a序列中找一个位置q,得以这个位置开始,以后每隔着aq+aq+1*(p)+.......aq+(m-1)*p,这个序列经过重新排序能够等于b,输出,所有的这样的位置. 题解:可以采用递推.先找起点是1,然后再找起点是1+p,找1+p的时候,其实只要在找1的基础上删除最前的那个数,然后再…

http://codeforces.com/contest/368/problem/D 题意:有a.b两个数组,a数组有n个数,b数组有m个数,现在给出一个p,要你找出所有的位置q,使得位置q  q+p   q+2*p  .....q+(m-1)*p   经过一定的操作(不改变数据大小)全等于b数组. 思路:首先肯定对a数组离散,然后二分对a.b数组分配好离散后的值.其实我们只需要枚举0————p位置,哈希记录,然后从q----一直滚到q(m-1)*p>=n,对于一个数据,出来第一个数,进去最后…

题目链接:http://codeforces.com/problemset/problem/314/C 题意:给定一个数列a.(1)写出a的不同的所有非下降子列:(2)定义某个子列的f值为数列中各个数的乘积.(3)求所有非下降子列的f值之和. 思路:我们用s[i]表示以数字a[i]结尾的所有非下降子列的f之和.那么a必然是接在之前小于等于a的某个数之后,设这个位置为j,那么s[i]=(s[j]+1)*a[i].也就是,a[i]可以接在其后或者自成一个子列.那么最后的答案就是所有的s值之和.这里有…

#include <bits/stdc++.h> #define llong long long using namespace std; const int N = 1e5; const int P = 1e9+; const llong INV2 = 5e8+; int nxt[N+]; int lstpos[N+]; int a[N+]; int num[N+]; int cnum[N+]; llong tmp0[N+],tmp1[N+],tmp2[N+]; vector<int&…

http://codeforces.com/contest/368/problem/C 从左向右记录从1位置到每一个位置上x,y,z的个数.然后判断在l,r区间内的x,y,z的关系满不满足abs(x-y)<=1&&abs(x-z)<=1&&abs(y-z)<=1,满足输出YES,否则输出NO. #include <cstdio> #include <cstring> #include <algorithm> #defin…

http://codeforces.com/contest/368/problem/B 从后往前找一遍就可以. #include <cstdio> #include <cstring> #include <algorithm> #define maxn 200000 using namespace std; int dp[maxn]; int n,m; int a[maxn]; bool vis[maxn]; int main() { while(scanf("…

D. Sereja ans Anagrams time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Sereja has two sequences a and b and number p. Sequence a consists of n integers a1, a2, ..., an. Similarly, sequence …