标签：tac style acm size 个数 scanf esc while mem
Problem L Tic-Tac-Toe
Problem DescriptionKim likes to play Tic-Tac-Toe.
Given a current state, and now Kim is going to take his next move. Please tell Kim if he can win the game in next 2 moves if both player are clever enough.
Here “next 2 moves” means Kim’s 2 move. (Kim move,opponent move, Kim move, stop).
Game rules:
Tic-tac-toe (also known as noughts and crosses or Xs and Os) is a paper-and-pencil game for two players, X and O, who take turns marking the spaces in a 3×3 grid. The player who succeeds in placing three of their marks in a horizontal, vertical, or diagonal row wins the game.
InputFirst line contains an integer T (1 ≤ T ≤ 10), represents there are T test cases.
For each test case: Each test case contains three lines, each line three string(“o” or “x” or “.”)(All lower case letters.)
x means here is a x
o means here is a o
. means here is a blank place.
Next line a string (“o” or “x”) means Kim is (“o” or “x”) and he is going to take his next move.
OutputFor each test case:
If Kim can win in 2 steps, output “Kim win!”
Otherwise output “Cannot win!”
Sample Input Sample Output
九宫棋Kim先下两步之内是否可以胜利
题解下次贴
#include <iostream> #include <stdio.h> #include <algorithm> #include <string.h> #include <cmath> using namespace std; char map[4][4]; int s[4][4],t,sum,k,cnt; bool judge(int x,int y){ if((x+y)%2==1){//当此时这个格子的行列的和为奇数时 cnt=0,sum=0; //cnt代表可能胜利的次数 for(int i=1;i<=3;i++){ sum+=s[i][y]; } sum*=k; //同是负数相乘为正数 if(sum==2) //空白数为2时代表可以 return true; else if(sum==1) //当sum和为1时，此时可能胜利 cnt++; sum=0; //注意，此时一定要重置sum，因为sum在这个函数的前面后面的含义不同 for(int i=1;i<=3;i++){ sum+=s[x][i];//计算此时画相同的符号的个数之和 } sum*=k; if(sum==2) //如果画相同符号的和为2，代表画下一个一定会胜利 return true; else if(sum==1) //如果此时画的相同的符合为1，代表可能胜利 cnt++; if(cnt==2){ //当可能胜利的次数超过2时一定可以胜利 return true; } } else { //当此时这个各自的行列的和为偶数时 cnt=0,sum=0; for(int i=1;i<=3;i++){ sum+=s[i][y]; } sum*=k; if(sum==2) return true; else if(sum==1) cnt++; sum=0; for(int i=1;i<=3;i++){ sum+=s[x][i]; } sum*=k; if(sum==2) return true; else if(sum==1) cnt++; sum=0; if(x==y){ //代表这个空白所在的地方在斜线上 for(int i=1;i<=3;i++){ sum+=s[i][i]; } sum*=k; if(sum==2) return true; else if(sum==1) cnt++; } else { for(int i=1;i<=3;i++){ sum+=s[i][4-i]; } sum*=k; if(sum==2) return true; else if(sum==1) cnt++; } if(cnt>=2){ return true; } } return false; } int main(){ char st; int flag; scanf("%d",&t); while(t--){ flag=0; memset(s,0,sizeof(0)); for(int i=1;i<=3;i++){ for(int j=1;j<=3;j++){ cin>>map[i][j]; if(map[i][j]==‘.‘) s[i][j]=0; else if(map[i][j]==‘o‘) s[i][j]=1; else if(map[i][j]==‘x‘) s[i][j]=-1; } } cin>>st; //代表此时Kim所用的符号 if(st==‘o‘) k=1; else if (st==‘x‘) k=-1; for(int i=1;i<=3;i++){ for(int j=1;j<=3;j++){ if(map[i][j]==‘.‘){ if(judge(i,j)) flag=1; } } } if(flag) puts("Kim win!"); else puts("Cannot win!"); } return 0; }
FZU Tic-Tac-Toe -.- FZU邀请赛 FZU 2283
标签：tac style acm size 个数 scanf esc while mem
原文地址：http://www.cnblogs.com/l609929321/p/7296104.html
