D. Iterated Linear Function 题目连接: http://www.codeforces.com/contest/678/problem/D Description Consider a linear function f(x) = Ax + B. Let's define g(0)(x) = x and g(n)(x) = f(g(n - 1)(x)) for n > 0. For the given integer values A, B, n and x find t…
  D. Iterated Linear Function time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Consider a linear function f(x) = Ax + B. Let's define g(0)(x) = x and g(n)(x) = f(g(n - 1)(x)) for n > 0. For…
D. Iterated Linear Function time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Consider a linear function f(x) = Ax + B. Let's define g(0)(x) = x and g(n)(x) = f(g(n - 1)(x))for n > 0. For the…
题目链接:http://codeforces.com/problemset/problem/678/D 简单的矩阵快速幂模版题 矩阵是这样的: #include <bits/stdc++.h> using namespace std; typedef __int64 LL; struct data { LL mat[][]; }; LL mod = 1e9 + ; data operator *(data a , data b) { data res; ; i <= ; ++i) { ;…
http://codeforces.com/contest/678/problem/D D. Iterated Linear Function Consider a linear function f(x) = Ax + B. Let's define g(0)(x) = x and g(n)(x) = f(g(n - 1)(x)) for n > 0. For the given integer values A, B, nand x find the value of g(n)(x) mod…
A. Johny Likes Numbers time limit per test 0.5 seconds memory limit per test 256 megabytes input standard input output standard output Johny likes numbers n and k very much. Now Johny wants to find the smallest integer x greater than n, so it is divi…
E. Another Sith Tournament 题目连接: http://www.codeforces.com/contest/678/problem/E Description The rules of Sith Tournament are well known to everyone. n Sith take part in the Tournament. The Tournament starts with the random choice of two Sith who wil…
C. Joty and Chocolate 题目连接: http://www.codeforces.com/contest/678/problem/C Description Little Joty has got a task to do. She has a line of n tiles indexed from 1 to n. She has to paint them in a strange pattern. An unpainted tile should be painted R…
B. The Same Calendar 题目连接: http://www.codeforces.com/contest/678/problem/B Description The girl Taylor has a beautiful calendar for the year y. In the calendar all days are given with their days of week: Monday, Tuesday, Wednesday, Thursday, Friday,…
A. Johny Likes Numbers 题目连接: http://www.codeforces.com/contest/678/problem/A Description Johny likes numbers n and k very much. Now Johny wants to find the smallest integer x greater than n, so it is divisible by the number k. Input The only line con…
http://codeforces.com/contest/678 A:水题 #include<bits/stdc++.h> #define fi first #define se second #define mp make_pair #define pb push_back #define pi acos(-1.0) #define ll long long #define mod 1000000007 #define C 0.5772156649 #define ls l,m,rt<…
Description Johny likes numbers n and k very much. Now Johny wants to find the smallest integer x greater than n, so it is divisible by the number k. Input The only line contains two integers n and k (1 ≤ n, k ≤ 109). Output Print the smallest intege…
题目链接: 题目 E. Another Sith Tournament time limit per test2.5 seconds memory limit per test256 megabytes inputstandard input outputstandard output 问题描述 The rules of Sith Tournament are well known to everyone. n Sith take part in the Tournament. The Tour…
数论题还是好恶心啊. 题目大意:给你两个不超过1e12的数 x,y,定义一个f ( x, y ) 如果y==0 返回 0 否则返回1+ f ( x , y - gcd( x , y ) ); 思路:我们设gcd ( x , y) 为G,那么 设 x  = A*G,y = B*G,我们考虑减去多少个G时x y 的gcd会改变,我们设减去 k个G的时候 x和y 的gcd为改变,即 A*G 和 ( B - k ) * G 的 gcd 改变了,什么情况下会改变呢,就是A 和( B -  k )的gcd…
Description Little Joty has got a task to do. She has a line of n tiles indexed from 1 to n. She has to paint them in a strange pattern. An unpainted tile should be painted Red if it's index is divisible by a and an unpainted tile should be painted B…
Description The girl Taylor has a beautiful calendar for the year y. In the calendar all days are given with their days of week: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday. The calendar is so beautiful that she wants to know wh…
/* CodeForces - 837E - Vasya's Function [ 数论 ] | Educational Codeforces Round 26 题意: f(a, 0) = 0; f(a, b) = 1 + f(a, b-gcd(a, b)); 求 f(a, b) , a,b <= 1e12 分析: b 每次减 gcd(a, b) 等价于 b/gcd(a,b) 每次减 1 减到什么时候呢,就是 b/gcd(a,b)-k 后 不与 a 互质 可先将 a 质因数分解,b能除就除,不能…
题目链接: D. Iterated Linear Function time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Consider a linear function f(x) = Ax + B. Let's define g(0)(x) = x and g(n)(x) = f(g(n - 1)(x)) for n > 0.…
Educational Codeforces Round 26 困到不行的场,等着中午显示器到了就可以美滋滋了 A. Text Volume time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You are given a text of single-space separated words, consisting of sm…
Educational Codeforces Round 17 A. k-th divisor 水题,把所有因子找出来排序然后找第\(k\)大 view code //#pragma GCC optimize("O3") //#pragma comment(linker, "/STACK:1024000000,1024000000") #include<bits/stdc++.h> using namespace std; function<voi…
Educational Codeforces Round 63 (Rated for Div. 2)题解 题目链接 A. Reverse a Substring 给出一个字符串,现在可以对这个字符串进行一次翻转,问是否存在一种方案,可以使得翻转后字符串的字典序可以变小.   这个很简单,贪心下就行了. 代码如下: Code #include <bits/stdc++.h> using namespace std; typedef long long ll; const int N = 3e5…
Educational Codeforces Round 39 (Rated for Div. 2) G 题意: 给一个序列\(a_i(1 <= a_i <= 10^{9}),2 <= n <= 200000\), 如果至多删除其中的一个数之后该序列为严格上升序列,那么称原序列为几乎严格上升序列. 现在每次将序列中的任意数字变成任意数字,问最少要操作几次才能将序列变成几乎严格上升子序列. 思路: 如果不考虑删除,求让整个序列都变成严格上升子序列的次数 求出\(序列a_i - i\)…
Educational Codeforces Round 41  D. Pair Of Lines 考虑先把凸包找出来,如果凸包上的点数大于\(4\)显然不存在解,小于等于\(2\)必然存在解 否则枚举凸包上两个点连线,判断剩余点能否被一条线覆盖即可 view code #pragma GCC optimize("O3") #pragma GCC optimize("Ofast,no-stack-protector") #include<bits/stdc++…
Educational Codeforces Round 39  D. Timetable 令\(dp[i][j]\)表示前\(i\)天逃课了\(j\)节课的情况下,在学校的最少时间 转移就是枚举第\(i\)天逃了\(x\)节课,然后取当天逃\(x\)节课情况下在学校的最小值即可 view code #pragma GCC optimize("O3") #pragma GCC optimize("Ofast,no-stack-protector") #include…
Educational Codeforces Round 30  A. Chores 把最大的换掉 view code #pragma GCC optimize("O3") #pragma GCC optimize("Ofast,no-stack-protector") #include<bits/stdc++.h> using namespace std; #define INF 0x3f3f3f3f #define endl "\n&quo…
Educational Codeforces Round 21  A. Lucky Year 个位数直接输出\(1\) 否则,假设\(n\)十进制最高位的值为\(s\),答案就是\(s-(n\mod s)\) view code #pragma GCC optimize("O3") #pragma GCC optimize("Ofast,no-stack-protector") #include<bits/stdc++.h> using namespac…
Educational Codeforces Round 43  A. Minimum Binary Number 显然可以把所有\(1\)合并成一个 注意没有\(1\)的情况 view code //#pragma GCC optimize("O3") //#pragma comment(linker, "/STACK:1024000000,1024000000") #include<bits/stdc++.h> using namespace std…
Educational Codeforces Round 20  A. Maximal Binary Matrix 直接从上到下从左到右填,注意只剩一个要填的位置的情况 view code //#pragma GCC optimize("O3") //#pragma comment(linker, "/STACK:1024000000,1024000000") #include<bits/stdc++.h> using namespace std; fu…
\(Educational\ Codeforces\ Round\ 85\ (Rated\ for\ Div.2)\) \(A. Level Statistics\) 每天都可能会有人玩游戏,同时一部分人会过关,玩游戏的人数和过关的人数会每天更新,问记录的数据是否没有矛盾 //#pragma GCC optimize("O3") //#pragma comment(linker, "/STACK:1024000000,1024000000") #include<…
Educational Codeforces Round 132 (Rated for Div. 2) A. Three Doors 简述 题意: 有三扇门(1~3), 其中两扇门后面有对应标号门的钥匙,现在手上有一把标号为n的钥匙,是否能打开所有的门? 判断现在有的钥匙 对应的门后 是否有钥匙即可,就是套娃 不是 Code #define OK cout << (ok ? "YES" : "NO") << endl void testcas…