ural 1932 The Secret of Identifier 容斥】的更多相关文章

主题链接:点击打开链接 stl+容斥 #include <cstdio> #include <cstring> #include <algorithm> #include <vector> #include <iostream> #include <set> using namespace std; #define N 65540 #define ll __int64 ll n; ll a[N][4], mul[4]={1,16,25…
Description Davy Jones: You've been captain of the Black Pearl for 13 years. That was our agreement. Jack: Technically I was only captain for two years, then I was mutinied upon. Davy Jones: Then you were a poor captain, but a captain nonetheless. Ha…
http://acm.timus.ru/problem.aspx?space=1&num=1932 B - The Secret of Identifier Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u Submit Status Practice URAL Description Davy Jones: You've been captain of the Black Pearl for 13 y…
标题效果: 计算到n字符串. 精确到只是有一个不同的字符,两个不同的字符.三个不同的字符,四对不同的字符. IDEAS: 枚举状态. dp[i] [j] ...当前串取出 i 状态下的全部字符转化成十进制数为 j 的出现的次数. 这种话,就记录了全部串的子串的状态. 然后计数就得到了全部的状态. 然后我们要得到精确不同的,能够用补集的思想,假设要精确到三个不同样,意味着要精确到1 个是同样的. 注意的问题是 在最后要运用容斥去重. #include <cstdio> #include <…
题意: 求1 - s 中 找出k个数 使它们的gcd  > 1 求这样的k个数的对数 解析: 从每个素数的倍数中取k个数  求方案数 然后素数组合,容斥一下重的 奇加偶减 莫比乌斯函数的直接套模板就好了 容斥函数为 mu[i] * -1 #include <iostream> #include <cstdio> #include <sstream> #include <cstring> #include <map> #include <…
Frogs Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5514 Description There are m stones lying on a circle, and n frogs are jumping over them.The stones are numbered from 0 to m−1 and the frogs are numbered fro…
Frogs Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1315    Accepted Submission(s): 443 Problem Description There are m stones lying on a circle, and n frogs are jumping over them.The stones a…
Frogs Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1509    Accepted Submission(s): 498 Problem Description  to m−1 and the frogs are numbered from 1 to n. The i-th frog can jump over exactly …
WLD is always very lucky.His secret is a lucky number . is a fixed odd number. Now he meets a stranger with numbers:.The stranger asks him questions.Each question is like this:Given two ranges and ,you can choose two numbers and to make .The you can…
  题意:规定每次跳的单位 a1, a2, a3 …… , an, M,次数可以为b1, b2, b3 …… bn, bn + 1, 正好表示往左,负号表示往右, 求能否调到左边一位,即 a1* b1+ a2 * b2 + a3 * b3 + …… + m * (bn + 1) = 1; 根据欧几里得,则a1, a2 a3 …… an, m 最大公约数为1,m已知且a1, a2, a3 …… an 均小于等于m, 一共有m ^ n可能, 将m 唯一分解之后, 假设m = 2 * 3 * 5, 则…