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…

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…

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…

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…

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…

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+;…