D. Soldier and Number Game time limit per test3 seconds memory limit per test256 megabytes inputstandard input outputstandard output Two soldiers are playing a game. At the beginning first of them chooses a positive integer n and gives it to the seco…

题目描述: Soldier and Number Game time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output Two soldiers are playing a game. At the beginning first of them chooses a positive integer n and gives it to t…

Description Two soldiers are playing a game. At the beginning first of them chooses a positive integer n and gives it to the second soldier. Then the second one tries to make maximum possible number of rounds. Each round consists of choosing a positi…

题目大意: 求  1(m)到n直接有多少个数字x满足 x可以整出这个数字的每一位上的数字 思路: 整除每一位.只需要整除每一位的lcm即可 但是数字太大,dp状态怎么表示呢 发现 1~9的LCM 是2520 ....也就是说只要对这个数mod2520 剩下的余数能整除lcm就可以整除了.. 计数的时候还有一个技巧,具体见注释 此外这个题还卡常数了,预处理lcm才过了.. 代码如下: #include <iostream> #include <stdio.h> #include<…

题目传送门 /* 题意:b+1,b+2,...,a 所有数的素数个数和 DP+埃氏筛法:dp[i] 记录i的素数个数和,若i是素数,则为1:否则它可以从一个数乘以素数递推过来 最后改为i之前所有素数个数和,那么ans = dp[a] - dp[b]: 详细解释:http://blog.csdn.net/catglory/article/details/45932593 */ #include <cstdio> #include <algorithm> #include <cs…

题目传送门 /* 题意:这题就是求b+1到a的因子个数和. 数学+DP:a[i]保存i的最小因子,dp[i] = dp[i/a[i]] +1;再来一个前缀和 */ /************************************************ Author :Running_Time Created Time :2015-8-1 14:08:34 File Name :B.cpp ************************************************…

D. Soldier and Number Game time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output Two soldiers are playing a game. At the beginning first of them chooses a positive integer n and gives it to the…

D. Soldier and Number Game Time Limit: 20 Sec  Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/546/problem/D Description Two soldiers are playing a game. At the beginning first of them chooses a positive integer n and gives it to the second s…

题目链接 输入t对数 a, b 求(b,a]内的每个数拆成素因子的个数和 这里每个数都可以写成素数的乘积,可以写成几个素数的和就有几个素因子,这里求的是(b,a]内的素因子和 思路: 素数的素因子个数是1 对于非素数A的素因子个数 = A/k  + 1 其中k是素数,也是第一个素数,或者K是比A小的数,并且A%k==0 下面是利用K是比A小的数,并且A%k==0 void solve(){ Scanner sc = new Scanner(System.in); int t = sc.nextI…

废话不多说,先上题目. 51nod Codeforces 两个其实是一个意思,看51nod题目就讲的很清楚了,题意不再赘述. 直接讲我的分析过程:刚开始拿到手有点蒙蔽,看起来很难,然后......然后我就开始一顿胡蒙,各种举例子.找规律下面为我取n = 10的过程. 1.首先1肯定不用取.因为它太特殊了,如果 取它的话,只能判断是不是1,因为所有数都是1的倍数:反之其他所有情况排除了就选它,不用浪费次数来取它. 2.素数肯定是要取的.因为......因为这题一看就是考素数的题. 3.以10为例,…