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: 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…
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 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…
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…
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…
题目链接: id=3370">http://poj.org/problem?id=3370 http://acm.hdu.edu.cn/showproblem.php?pid=1808 题目大意: 给你两个整数C和N,再给你N个正数的序列.从中找到若干数,使得其和刚好是 C 的倍数. 输出这些数的序号. 解题思路: 典型的抽屉原理. Sum[i]为序列中前 i 项的和.则有两种可能: 1.若有 Sum[i] 是 C 的倍数.则直接输出前 i 项. 2.假设没有不论什么的 Sum[i] 是…
[POJ3370]&[HDU1808]Halloween treats 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…