基本原理:n+1只鸽子飞回n个鸽笼至少有一个鸽笼含有不少于2只的鸽子. 很简单,应用却也很多,很巧妙,看例题: Description Every year there is the same problem at Halloween: Each neighbour is only willing to give a certain total number of sweets on that day, no matter how many children call on him, so it…
Halloween treats Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 7644   Accepted: 2798   Special Judge Description Every year there is the same problem at Halloween: Each neighbour is only willing to give a certain total number of sweets…
链接:传送门 题意:万圣节到了,有 c 个小朋友向 n 个住户要糖果,根据以往的经验,第i个住户会给他们a[ i ]颗糖果,但是为了和谐起见,小朋友们决定要来的糖果要能平分,所以他们只会选择一部分住户索要糖果,这样糖果恰好可以平分又不会剩下,输出索要糖果的用户编号.如果没有任何一组住户给的糖果总数能够平分,则输出 "no sweets" .SPJ( 意味着答案不唯一 ) 思路: 简单鸽巢原理题目,可以将问题转化成 给出一个正整数序列A1,A2, ... ,An ,求有没有整数 l 和…
Halloween treats 和POJ2356差点儿相同. 事实上这种数列能够有非常多,也能够有不连续的,只是利用鸽巢原理就是方便找到了连续的数列.并且有这种数列也必然能够找到. #include <cstdio> #include <cstdlib> #include <xutility> int main() { int c, n; while (scanf("%d %d", &c, &n) && c) { i…
Halloween treats Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 7143   Accepted: 2641   Special Judge Description Every year there is the same problem at Halloween: Each neighbour is only willing to give a certain total number of sweets…
Halloween treats Every year there is the same problem at Halloween: Each neighbour is only willing to give a certain total number of sweets on that day, no matter how many children call on him, so it may happen that a child will get nothing if it is…
Halloween treats Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 6631   Accepted: 2448   Special Judge Description Every year there is the same problem at Halloween: Each neighbour is only willing to give a certain total number of sweets…
Description Every year there is the same problem at Halloween: Each neighbour is only willing to give a certain total number of sweets on that day, no matter how many children call on him, so it may happen that a child will get nothing if it is too l…
POJ 2356: 题目大意: 给定n个数,希望在这n个数中找到一些数的和是n的倍数,输出任意一种数的序列,找不到则输出0 这里首先要确定这道题的解是必然存在的 利用一个 sum[i]保存前 i 个数的和对n的取模 sum[0] = 0; 那么sum[0] ~ sum[n]有n+1个数据,这些数据的范围都是 0~n , 要是存在 sum[i] = 0,那么输出前 i 个数据即可 要是不存在那根据鸽巢原理可以说明必然能找到一个 sum[i] = sum[j]  ,那么说明 (sum[i+1] +…
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…