A. Shell Game
A. Shell Game time limit per test 0.5 seconds memory limit per test 256 megabytes input standard input output standard output Bomboslav likes to look out of the window in his room and watch lads outside playing famous shell game. The game is played by two persons: operator and player. Operator takes three similar opaque shells and places a ball beneath one of them. Then he shuffles the shells by swapping some pairs and the player has to guess the current position of the ball. Bomboslav noticed that guys are not very inventive,so the operator always swaps the left shell with the middle one during odd moves (first,third,fifth,etc.) and always swaps the middle shell with the right one during even moves (second,fourth,etc.). Let's number shells from0to2from left to right. Thus the left shell is assigned number0,the middle shell is1and the right shell is2. Bomboslav has missed the moment when the ball was placed beneath the shell,but he knows that exactlynmovements were made by the operator and the ball was under shellxat the end. Now he wonders,what was the initial position of the ball? Input The first line of the input contains an integern(1?≤?n?≤?2·109)— the number of movements made by the operator. The second line contains a single integerx(0?≤?x?≤?2)— the index of the shell where the ball was found afternmovements. Output Print one integer from0to2— the index of the shell where the ball was initially placed. Examples input 4 2 output 1 input 1 1 output 0 Note In the first sample,the ball was initially placed beneath the middle shell and the operator completed four movements.
——————————————————————————————————— 题目的意思是给你3个杯子,其中一个有物品,每次奇数次交换0 1 两个,偶数次交换1 2两个,给出操作次数和最后物品的位置,求出物品起始位置 思路:找出6循环节,mod之后反向推 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; int main() { int n,m; while(~scanf("%d%d",&n,&m)) { n%=6; if(m==0) { if(n==1||n==2) printf("1n"); else if(n==3||n==4) printf("2n"); else printf("0n"); } else if(m==1) { if(n==0||n==3) printf("1n"); else if(n==2||n==5) printf("2n"); else printf("0n"); } else { if(n==4||n==5) printf("1n"); else if(n==0||n==1) printf("2n"); else printf("0n"); } } return 0; } (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |