HDU 1034 分糖果 (模拟题)
来源:互联网 发布:网络超市好开吗 编辑:程序博客网 时间:2024/05/16 12:59
题目说的是一个分糖果的游戏,n个学生围成一圈,每个人手上有a[i]个糖果,每一轮游戏开始时,学生手中的糖果都是偶数,他们都把自己手中一半的糖果分给他右边的同学。分完后手中糖果数是奇数的,老师会给他一颗糖果。如果所有人手中的糖果数都一样 游戏结束。
输出总共进行了多少轮游戏 还有最后学生手中的糖果数。
Problem Description
A number of students sit in a circle facing their teacher in the center. Each student initially has an even number of pieces of candy. When the teacher blows a whistle, each student simultaneously gives half of his or her candy to the neighbor on the right. Any student, who ends up with an odd number of pieces of candy, is given another piece by the teacher. The game ends when all students have the same number of pieces of candy.
Write a program which determines the number of times the teacher blows the whistle and the final number of pieces of candy for each student from the amount of candy each child starts with.
Write a program which determines the number of times the teacher blows the whistle and the final number of pieces of candy for each student from the amount of candy each child starts with.
Input
The input may describe more than one game. For each game, the input begins with the number N of students, followed by N (even) candy counts for the children counter-clockwise around the circle. The input ends with a student count of 0. Each input number is on a line by itself.
Output
For each game, output the number of rounds of the game followed by the amount of candy each child ends up with, both on one line.
Sample Input
6362222211222018161412108642424680
Sample Output
15 1417 224 8HintThe game ends in a finite number of steps because:1. The maximum candy count can never increase.2. The minimum candy count can never decrease.3. No one with more than the minimum amount will ever decrease to the minimum.4. If the maximum and minimum candy count are not the same, at least one student with the minimum amount must have their count increase.
#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int a[1000];int main(){int num,n,i,j,t,last;bool up;while(scanf("%d",&n)==1 && n) {for(i=1;i<=n;i++) scanf("%d",&a[i]);num=0;up=true;while(up) {up=false;last=a[1]/2;for(i=2;i<=n;i++) {t=a[i];a[i]=a[i]/2+last;if(a[i]&1) a[i]++;last=t/2;}a[1]=a[1]/2+last;if(a[1]&1) a[1]++;for(i=2;i<=n;i++) {if(a[i]!=a[i-1]) {up=true;break;}}num++;}printf("%d %d\n",num,a[1]);}return 0;}
0 0
- HDU 1034 分糖果 (模拟题)
- 分糖果(模拟)
- 蓝桥杯 - 分糖果 (模拟~)
- 蓝桥杯--历届试题 分糖果(模拟水题)
- 蓝桥杯 历届试题 分糖果 (模拟)
- 蓝桥 分糖果 (模拟)
- 历届试题 分糖果(模拟)
- HPU1290 分糖果 【模拟】
- 蓝桥杯 分糖果(模拟)
- 【模拟试题】分糖果
- 历届试题 分糖果 【模拟】
- PREV-32分糖果(模拟)
- 历届试题 分糖果 模拟
- 蓝桥杯 历届试题 分糖果 (简单模拟)
- 蓝桥杯历届试题——分糖果(模拟)
- 算法题:分糖果
- 分糖果 (蓝桥杯)
- 蓝桥杯 历届试题 分糖果 模拟
- Exception raised during rendering: java.lang.System.arraycopy([CI[CII)V Exception details are logged
- 1038 统计同成绩学生PAT
- Java并发编程:volatile关键字解析
- cookie的弊端
- apt-get命令
- HDU 1034 分糖果 (模拟题)
- js----Date
- 创建React Native 离线 APP的过程
- 登陆拦截 filter() , 字符编码拦截器
- 正则表达式-前后查找
- 出现“MSCOMCTL.OCX或其附件之一不能正确使用:一个文件丢失或无效或'comctl32.ocx'其附件之一不能正确使用:一个文件丢失或无效”的提示的解决办法
- mysql insert操作
- 3GPP 36211-c70-8(内容不太懂)
- 多少个1?(位运算)