[POJ2356]Find a multiple Description -The input contains N natural (i.e. positive integer) numbers ( N <= 10000 ). Each of that numbers is not greater than 15000. This numbers are not necessarily different (so it may happen that two or more of them w…

/* POJ-2356 Find a multiple ----抽屉原理 Find a multiple Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 4228 Accepted: 1850 Special Judge Description The input contains N natural (i.e. positive integer) numbers ( N <= 10000 ). Each of that nu…

Find a multiple Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 7133 Accepted: 3122 Special Judge Description The input contains N natural (i.e. positive integer) numbers ( N <= 10000 ). Each of that numbers is not greater than 15000. This…

Find a multiple Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 8776   Accepted: 3791   Special Judge Description The input contains N natural (i.e. positive integer) numbers ( N <= 10000 ). Each of that numbers is not greater than 15000…

/* 引用过来的 题意: 给出N个数,问其中是否存在M个数使其满足M个数的和是N的倍数,如果有多组解, 随意输出一组即可.若不存在,输出 0. 题解: 首先必须声明的一点是本题是一定是有解的.原理根据抽屉原理: 因为有n个数,对n个数取余,如果余数中没有出现0,根据鸽巢原理,一定有两个数的余数相同, 如果余数出现0,自然就是n的倍数.也就是说,n个数中一定存在一些数的和是n的倍数. 本题的思路是从第一个数开始一次求得前 i(i <= N)项的和关于N的余数sum,并依次记录相应余数的存在状态,…

题意:给你N个数,从中取出任意个数的数 使得他们的和 是 N的倍数: 在鸽巢原理的介绍里面,有例题介绍:设a1,a2,a3,……am是正整数的序列,试证明至少存在正数k和l,1<=k<=l<=m,是的和ak+ak+1+……+al是m的倍数,接下来开始证明: 构造一个序列s1=a1,s2=a1+a2,……,sm=a1+a2+……+am,那么会产生两种可能: 1:若有一个sn是m的倍数,那么定理成立: 2:假设上述的序列中没有任何一个元素是m的倍数,令rh ≡ sh mod m;其中h=1,…

题意: 给定n个数,从中选取m个数,使得\(\sum | n\).本题使用Special Judge. 题解: 既然使用special judge,我们可以直接构造答案. 首先构造在mod N剩余系下的前缀和. \[sum_i = (a_i + sum_{i-1}) mod n\] 剩余系N的完系中显然共有N-1个元素,我们有N个前缀和. 根据鸽巢原理,一定有\(sum_j = sum_i\) 所以这样构造是可行的. TRICK 具体实现的时候用了一个技巧: 从前往后扫描sum数组,记录一个po…

Conversion to Dalvik format failed: Unable to execute dex: Multiple dex files define ... 这个错误是因为有两个相同的jar包,删除其中一个就可以正常运行了.…

Find a multiple Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7192   Accepted: 3138   Special Judge Description The input contains N natural (i.e. positive integer) numbers ( N <= 10000 ). Each of that numbers is not greater than 15000…

The API: int read4(char *buf) reads 4 characters at a time from a file. The return value is the actual number of characters read. For example, it returns 3 if there is only 3 characters left in the file. By using the read4 API, implement the function…