uva 11489Integer Game
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原题:
Two players, S and T, are playing a game where they make alternate moves. S plays first.
In this game, they start with an integer N. In each move, a player removes one digit from the
integer and passes the resulting number to the other player. The game continues in this fashion until a player finds he/she has no digit to remove when that player is declared as the loser.
With this restriction, its obvious that if the number of digits in N is odd then S wins otherwise T
wins. To make the game more interesting, we apply one additional constraint. A player can remove a particular digit if the sum of digits of the resulting number is a multiple of 3 or there are no digits left. Suppose N = 1234. S has 4 possible moves. That is, he can remove 1, 2, 3, or 4. Of these, two of
them are valid moves.
• Removal of 4 results in 123 and the sum of digits = 1 + 2 + 3 = 6; 6 is a multiple of 3.
• Removal of 1 results in 234 and the sum of digits = 2 + 3 + 4 = 9; 9 is a multiple of 3.
The other two moves are invalid.
If both players play perfectly, who wins?
Input
The first line of input is an integer T (T < 60) that determines the number of test cases. Each case is
a line that contains a positive integer N. N has at most 1000 digits and does not contain any zeros.
Output
For each case, output the case number starting from 1. If S wins then output ‘S’ otherwise output ‘T’.
Sample Input
3
4
33
771
Sample Output
Case 1: S
Case 2: T
Case 3: T
中文:
给你一个1000位数,现在两个人轮流从这些数中取走一个数字,要求拿走后剩下的数字可以被3整除。不能取数的人判定输。先手胜输出S后手胜输出T。(注意,如果给的初始数据不能被3整除,不能直接判定后手胜!)
#include <bits/stdc++.h>using namespace std;//fstream in,out;int main(){ ios::sync_with_stdio(false); string s; int mark3,tot,mark2,mark1; int t,k=1;// in.open("data.txt");// out.open("answer.txt"); cin>>t; while(t--) { cin>>s; tot=mark3=mark2=mark1=0; for(int i=0;i<s.size();i++) { tot+=s[i]-'0'; if(s[i]=='0'||s[i]=='3'||s[i]=='6'||s[i]=='9') mark3++; if(s[i]=='1'||s[i]=='4'||s[i]=='7') mark1++; if(s[i]=='2'||s[i]=='5'||s[i]=='8') mark2++; } cout<<"Case "<<k++<<": "; if(s.size()==1) { cout<<'S'<<endl; continue; } if(tot%3==0) { if(mark3%2) cout<<'S'<<endl; else cout<<'T'<<endl; } else { if(tot%3==1) { if(mark1==0) cout<<'T'<<endl; else { if(mark3%2) cout<<'T'<<endl; else cout<<'S'<<endl; } } else { if(mark2==0) cout<<'T'<<endl; else { if(mark3%2) cout<<'T'<<endl; else cout<<'S'<<endl; } } } }// in.close();// out.close(); return 0;}
解答:
简单的博弈问题
首先要知道一个小知识点就是如果一个数的各位的和能被3整除,那么这个数就能被3整除。
两个人轮流取数,首先判断这个数能不能被3整除,如果能。那么判断这个数里面有多少个3的倍数,因为在这个数是3的倍数的前提下,只有每次取出3的倍数,剩下的数的和才能是3的倍数。
如果这个数不是3的倍数,那么就要先取出一个和这个数同余的数,比如890899这个数的和是43,模上3为1,那么就要取出1,4,7这三个数之一,如果没有就是先手负,如果有这个数,取出一个此数,那么剩下的数的和就一定能被3正常,继续用3的倍数的个数去判断即可,不过,先手已经取过数了,答案要反着写。
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