版权声明:本文为博主原创文章.未经博主同意不得转载. https://blog.csdn.net/u011328934/article/details/37612503 题目链接:uva 11237 - Halloween treats 题目大意:有c个小孩要到邻居家去要糖果.有n户邻居.每户邻居仅仅会提供固定数量的糖果,熊孩子们为了不发生冲突,决定将取来的糖果平均分配,问说取那几家邻居的糖果能够做到平均分配.注意n ≥ c. 解题思路:抽屉原理.求出序列的前缀和,有n个,将前缀和对c取模后.依…

11237 - Halloween treats option=com_onlinejudge&Itemid=8&page=show_problem&category=516&problem=2178&mosmsg=Submission+received+with+ID+13856428" target="_blank" style="">题目链接 题意:有c个小伙伴,n个房子(c <= n).每一个房子…

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: 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…

Repeating Decimals 紫书第3章,这哪是模拟啊,这是数论题啊 [题目链接]Repeating Decimals [题目类型]抽屉原理 &题解: n除以m的余数只能是0~m-1,根据抽屉原则,当计算m+1次时至少存在一个余数相同,即为循环节:存储余数和除数,输出即可. 上面是我查到的,现在让我解释一下: 比如5/43,先是要模拟除法运算,第一步50/43 余7;第二步,70/43 于27... 这样一直取余下去,肯定不会超过43次,就会有余数相同的情况,有相同情况就是找到循环节了.…

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…

题目传送:http://acm.hdu.edu.cn/showproblem.php?pid=1808 Problem 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, s…

Halloween treats Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1097    Accepted Submission(s): 435Special Judge Problem Description Every year there is the same problem at Halloween: Each neig…

[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…

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…