POJ - 1978 Hanafuda Shuffle
最初给牌编号时,编号的顺序是从下到上;洗牌时,认牌的顺序是从上到下。注意使用循环是尽量统一“i”的初始化值,都为“0”或者都为“1”,限界条件统一使用“<”或者“<=”。
POJ - 1978 Hanafuda Shuffle
Time Limit: 1000MS Memory Limit: 30000KB 64bit IO Format: %I64d & %I64u
Description
There are a number of ways to shuffle a deck of cards. Hanafuda shuffling for Japanese card game 'Hanafuda' is one such example. The following is how to perform Hanafuda shuffling.
There is a deck of n cards. Starting from the p-th card from the top of the deck, c cards are pulled out and put on the top of the deck, as shown in Figure 1. This operation, called a cutting operation, is repeated.
Write a program that simulates Hanafuda shuffling and answers which card will be finally placed on the top of the deck.
Figure 1: Cutting operation
Input
The input consists of multiple data sets. Each data set starts with a line containing two positive integers n (1 <= n <= 50) and r (1 <= r <= 50); n and r are the number of cards in the deck and the number of cutting operations, respectively.
There are r more lines in the data set, each of which represents a cutting operation. These cutting operations are performed in the listed order. Each line contains two positive integers p and c (p + c <= n + 1). Starting from the p-th card from the top of the deck, c cards should be pulled out and put on the top.
The end of the input is indicated by a line which contains two zeros.
Each input line contains exactly two integers separated by a space character. There are no other characters in the line.
Output
For each data set in the input, your program should write the number of the top card after the shuffle. Assume that at the beginning the cards are numbered from 1 to n, from the bottom to the top. Each number should be written in a separate line without any superfluous characters such as leading or following spaces.
Sample Input
5 2
3 1
3 1
10 3
1 10
10 1
8 3
0 0
Sample Output
4
4
Source
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h> struct card{
int i;
}; card cards[];
card t[]; int main()
{
int n = , r = , p = , c = ;
while(scanf("%d%d", &n, &r)) {
if(!n && !r) break;
for(int i = ; i <= n; i++) {
cards[i].i = n-i+;
}
for(int i = ; i <= r; i++) {
scanf("%d%d", &p, &c);
for(int j = ; j <= c; j++) { //抽牌
t[j] = cards[p+j-];
}
for(int j = p-; j >= ; j--) { //顶牌下落
cards[j+c] = cards[j];
}
for(int j = ; j <= c; j++) { //放顶牌
cards[j] = t[j];
}
}
printf("%d\n", cards[].i);
} return ;
}
POJ - 1978 Hanafuda Shuffle的更多相关文章
- POJ 3590 The shuffle Problem
Any case of shuffling of n cards can be described with a permutation of 1 to n. Thus there are total ...
- poj 3590 The shuffle Problem——DP+置换
题目:http://poj.org/problem?id=3590 bzoj 1025 的弱化版.大概一样的 dp . 输出方案的时候小的环靠前.不用担心 dp 时用 > 还是 >= 来转 ...
- 【POJ - 3087】Shuffle'm Up(模拟)
Shuffle'm Up 直接写中文了 Descriptions: 给定两个长度为len的字符串s1和s2, 接着给出一个长度为len*2的字符串s12. 将字符串s1和s2通过一定的变换变成s12, ...
- POJ 3087:Shuffle'm Up
Shuffle'm Up Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 7364 Accepted: 3408 Desc ...
- POJ 1978
#include <iostream> #define MAXN 55 using namespace std; int _m[MAXN]; int tem[MAXN]; void cop ...
- POJ 3590 The shuffle Problem [置换群 DP]
传送门 $1A$太爽了 从此$Candy?$完全理解了这种$DP$做法 和bzoj1025类似,不过是求最大的公倍数,并输出一个字典序最小的方案 依旧枚举质因子和次数,不足的划分成1 输出方案从循环长 ...
- POJ 题目分类(转载)
Log 2016-3-21 网上找的POJ分类,来源已经不清楚了.百度能百度到一大把.贴一份在博客上,鞭策自己刷题,不能偷懒!! 初期: 一.基本算法: (1)枚举. (poj1753,poj2965 ...
- (转)POJ题目分类
初期:一.基本算法: (1)枚举. (poj1753,poj2965) (2)贪心(poj1328,poj2109,poj2586) (3)递归和分治法. (4)递推. ...
- poj分类
初期: 一.基本算法: (1)枚举. (poj1753,poj2965) (2)贪心(poj1328,poj2109,poj2586) (3)递归和分治法. ( ...
随机推荐
- PLSQL Developer注册码
Product Code:4t46t6vydkvsxekkvf3fjnpzy5wbuhphqzserial Number:601769password:xs374ca
- linux中,常用的账号管理命令
创建新用户:adduser 用户名创建新用户并将其加入一个现有组中:adduser 用户名 -G 组名创建新用户并使其只属于该组:adduser 用户名 -g 组名创建用户密码:passwd 用户名创 ...
- HDU2955 背包DP
Robberies Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total S ...
- JSP + AJAX完整实例及代码
(1)发送请求index.jsp,注意引入jquery.min.js文件 <%@ page language="java" contentType="text/ht ...
- centos下安装mysql遇到的问题
我前边遇到的坎,竟然和这篇百度经验惊人的相似,他帮我大忙了,十分感谢作者啊,连接双手奉上http://jingyan.baidu.com/article/ce436649fec8533773afd38 ...
- UIDynamic(物理仿真)
简介 什么是UIDynamic UIDynamic是从iOS 7开始引入的一种新技术,隶属于UIKit框架 可以认为是一种物理引擎,能模拟和仿真现实生活中的物理现象 如: 重力.弹性碰撞等现象 物理引 ...
- 获取Java系统相关信息
package com.test; import java.util.Properties; import java.util.Map.Entry; import org.junit.Test; pu ...
- # 20145334赵文豪 《Java程序设计》第6周学习总结
20145334赵文豪 <Java程序设计>第6周学习总结 教材学习内容总结 第十章 输入/输出 数据流 I/O操作主要是指使用Java进行输入,输出操作. Java所有的I/O机制都是基 ...
- ListView的深入学习
ListView通常有两个职责: 将数据填充到布局 : 处理用户的点击选择操作 二.创建ListView需要3个元素 ListView的每一列的View View的数据或者图片 连接数据与ListVi ...
- IOS第七天(5:UiTableView 汽车品牌,复杂模型分组展示,A-Z索要列表) (2015-08-05 14:03)
复杂模型分组展示 #import "HMViewController.h" #import "HMCarGroup.h" #import "HMCar ...