A. Grid game

// Problem: Shopping// Contest: NowCoder// URL: Memory Limit: 1048576 MB// Time Limit: 2000 ms// 2022-03-04 19:55:28// // Powered by CP Editor (namespace std;#define rep(i,l,r) for(int i=(l);i<=(r);i++)#define per(i,l,r) for(int i=(l);i>=(r);i--)#define ll long long#define pii pair#define mset(s,t) memset(s,t,sizeof(t))#define mcpy(s,t) memcpy(s,t,sizeof(t))#define fi first#define se second#define pb push_back#define all(x) (x).begin(),(x).end()#define SZ(x) ((int)(x).size())#define mp make_pairconst ll mod = 1e9 + 7;inline ll qmi (ll a, ll b) { ll ans = 1; while (b) { if (b & 1) ans = ans * a%mod; a = a * a %mod; b >>= 1; } return ans;}inline int read () { int x = 0, f = 0; char ch = getchar(); while (!isdigit(ch)) f |= (ch=='-'),ch= getchar(); while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar(); return f?-x:x;}template void print(T x) { if (x < 0) putchar('-'), x = -x; if (x >= 10) print(x/10); putchar(x % 10 + '0');}inline ll sub (ll a, ll b) { return ((a - b ) %mod + mod) %mod;}inline ll add (ll a, ll b) { return (a + b) %mod;}inline ll inv (ll a) { return qmi(a, mod - 2);}void solve() { string s; cin >> s; int a = 0, b = 0; for (auto t : s) { if (t == '0') { if (a) { cout << "3 4\n"; } else { cout <<"1 4\n"; } a = !a; } else { if (b) { cout << "4 3" << endl; } else cout << "4 1" << endl; b = !b; } }}int main () { int t; t =1; //cin >> t; while (t --) solve(); return 0;}

不记得题目已知条i安乐。已知4*4的方块 我们可以从消除条件来每次连续两个必定消除。这样就不会相交

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