Codeforces Round #257 (Div. 2) A. Jzzhu and Children解题报告
来源:互联网 发布:搭建php开发环境 编辑:程序博客网 时间:2024/06/05 18:00
对于这题笔者认为这符合队列的先进先出原则,所以可以通过队列数据结构来实现这个问题,
至于序号的话就可以通过建立结构体来达到要求;
而对于数据结构队列笔者在这稍微提示一下
像栈一样,队列(queue)也是一种线性表,它的特性是先进先出,插入在一端,删除在另一端。就像排队一样,刚来的人入队(push)要排在队尾(rear),每次出队(pop)的都是队首(front)的人。如图1,描述了一个队列模型。
下面就贴出笔者的代码:
#include<cstdio>#include<queue>#define N 100+10#include<algorithm>using namespace std;struct child{ int need; int number;};int main(void){ struct child Child[N],p,temp; queue<child> q; int n,m; scanf("%d%d",&n,&m); for(int i=1;i<=n;++i) Child[i].number = i; for(int i=1;i<=n;++i){ scanf("%d",&Child[i].need); q.push(Child[i]); } if(q.size()==1) { printf("1\n"); return 0;} while(q.size()!=1) { p = q.front(); if(p.need > m) { p.need -= m; temp.need = p.need; temp.number = p.number; q.pop(); q.push(temp); } else q.pop(); } p = q.front(); printf("%d\n",p.number); return 0;}
更多关于队列的知识可以点击下面链接:
点击打开链接
0 0
- Codeforces Round #257 (Div. 2) A. Jzzhu and Children解题报告
- Codeforces Round #257 (Div. 2) A. Jzzhu and Children
- Codeforces Round #257 (Div. 2)1A Jzzhu and Children
- Codeforces Round #257 (Div. 2) A. Jzzhu and Children
- Codeforces Round #257 (Div. 2)A. Jzzhu and Children
- Codeforces Round #257 (Div. 2) A. Jzzhu and Children
- Codeforces Round #257 (Div. 2) A. Jzzhu and Children
- Codeforces Round #257 (Div. 2) Jzzhu and Children (队列)
- Codeforces Round #257 (Div. 2) 450A - Jzzhu and Children(模拟)
- Codeforces Round #257 (Div. 2) A. Jzzhu and Children(简单题)
- Codeforces Div. 2 #257-A. Jzzhu and Children
- Codeforces #257 div 2 A. Jzzhu and Children
- Codeforces Round #257 (Div. 2)449A - Jzzhu and Chocolate
- Codeforces Round #257 (Div. 2/A)/Codeforces450A_Jzzhu and Children
- cf #257(Div.2) A. Jzzhu and Children
- Codeforces Round #257 (Div. 2) B. Jzzhu and Sequences
- Codeforces Round #257 (Div. 2) B. Jzzhu and Sequences
- Codeforces Round #257 (Div. 2) B. Jzzhu and Sequences
- C++静态类
- 数字统计
- 杭电ACM3549(最大流)
- Qemu+Gdb debug linux kernel
- C语言数组和指针
- Codeforces Round #257 (Div. 2) A. Jzzhu and Children解题报告
- Strace 工具
- mysql5.7修改root密码
- 实战c++中的vector系列--知道emplace_back为何优于push_back吗?
- 一次mysql瘫痪解救
- [从头学数学] 第07节 11~20各数的认识
- 51nod 1422:沙拉酱前缀
- 数列求和
- 媒体服务器性能评估必须考虑的几个指标