https://ac.nowcoder.com/acm/contest/338/I
题解:首先轮到出手的时候如果在(0,0)上肯定是输的,而(0,1)(1,0)(0,2)(2,0)(1,1)肯定是赢的;
往上递推,某一个(x,y)如果可以走的(x-1,y)(x,y-1)(x-1,y-1)三点都是必输的,那么在(x,y)的人必输。
借大佬代码一用
#include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int using namespace std; typedef long long ll; //typedef __int128 lll; const int N=1000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; int t,n,m,k,q,ans; bool a[N][N]; char str; int main() { #ifdef DEBUG freopen("input.in", "r", stdin); //freopen("output.out", "w", stdout); #endif scanf("%d%d",&n,&m); a[0][0]=0; a[1][0]=a[0][1]=1; a[1][1]=1; for(int i=0; i<=n; i++) { for(int j=0; j<=m; j++) { if(i-1>=0&&j-1>=0) { if(a[i][j-1]&&a[i-1][j]&&a[i-1][j-1]) a[i][j]=0; else a[i][j]=1; } else if(i-1>=0) { if(a[i-1][j]) a[i][j]=0; else a[i][j]=1; } else if(j-1>=0) { if(a[i][j-1]) a[i][j]=0; else a[i][j]=1; } } } if(a[n][m]) { cout << "ii" << endl; } else { cout << "gg" << endl; } //cout << "Hello world!" << endl; return 0; }
待补充
#include<bits/stdc++.h> using namespace std; int main(){ int m,n; cin >> m >>n; if(min(m,n)%2) cout <<"ii"; else if ((max(m,n)-min(m,n))%2==0) cout <<"gg"; else cout <<"ii"; return 0; }
原文地址:https://www.cnblogs.com/DWVictor/p/10229965.html
时间: 2024-11-05 02:04:01