希望一年更比一年水
Secret Chamber at Mount Rushmore
签到题，搞出可达矩阵就可以了
1 #include <bits/stdc++.h> 2 using namespace std; 3 int d[26][26]; 4 5 int main() { 6 int m, n; 7 scanf("%d %d", &m, &n); 8 for(int i = 0; i < 26; ++i) d[i][i] = 1; 9 for(int i = 1; i <= m; ++i) { 10 char a[11], b[11]; 11 scanf("%s %s", a, b); 12 d[(a[0] - 'a')][(b[0] - 'a')] = 1; 13 } 14 for(int i = 0; i < 26; ++i) 15 for(int j = 0; j < 26; ++j) 16 for(int k = 0; k < 26; ++k) 17 d[j][k] |= d[j][i] & d[i][k]; 18 for(int i = 1; i <= n; ++i) { 19 char a[111], b[111]; 20 scanf("%s %s", a, b); 21 int len1 = strlen(a), len2 = strlen(b), ok = 1; 22 if(len1 != len2) ok = 0; 23 else for(int j = 0; j < len1; ++j) { 24 if(!d[a[j] - 'a'][b[j] - 'a']) { 25 ok = 0; break; 26 } 27 } 28 puts(ok ? "yes" : "no"); 29 } 30 return 0; 31 }
又一签到题，二分答案判断总时间是否相等，注意速度不能变负
1 #include <bits/stdc++.h> 2 using namespace std; 3 const double eps = 1e-8; 4 int d[1111], s[1111]; 5 6 int main() { 7 int n, t; 8 double l = -1e18, r = 1e18; 9 scanf("%d %d", &n, &t); 10 for(int i = 1; i <= n; ++i) { 11 scanf("%d %d", d + i, s + i); 12 l = max(l, -1.0 * s[i]); 13 } 14 while(r - l > eps) { 15 double m = (l + r) / 2; 16 double sumt = 0; 17 for(int i = 1; i <= n; ++i) 18 sumt += d[i] / (s[i] + m); 19 if(sumt > t) l = m; 20 else r = m; 21 } 22 printf("%.8f ", l); 23 return 0; 24 }
好像比较正常是一个n3的dp但是我失去理智n4就过了
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 int r[266], p[266], id[266]; 5 ll f[266][266], g[266][266]; 6 const ll inf = 1e18; 7 8 bool cmp(int i, int j) { 9 return r[i] < r[j]; 10 } 11 12 int main() { 13 int d, k; 14 scanf("%d %d", &d, &k); 15 for(int i = 1; i <= d; ++i) { 16 scanf("%d %d", r + i, p + i); 17 id[i] = i; 18 } 19 sort(id + 1, id + 1 + d, cmp); 20 for(int i = 0; i < 266; ++i) 21 for(int j = 0; j < 266; ++j) 22 f[i][j] = inf; 23 f[0][0] = 0; 24 for(int i = 0; i <= 255; ++i) { 25 memcpy(g, f, sizeof(g)); 26 for(int j = 0; j <= d; ++j) { 27 for(int l = 0; l < k; ++l) { 28 if(g[j][l] == inf) continue; 29 ll sum = 0; 30 for(int q = 1; j + q <= d; ++q) { 31 int x = id[j + q]; 32 sum += 1ll * p[x] * (r[x] - i) * (r[x] - i); 33 f[j + q][l + 1] = min(f[j + q][l + 1], g[j][l] + sum); 34 } 35 } 36 } 37 } 38 printf("%lld", f[d][k]); 39 return 0; 40 }
好像最大匹配就可以了但是我失去理智最大权匹配就过了
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 int a[111][111], rmax[111], cmax[111]; 5 6 // SPFA_min_cost_flow 7 const ll inf = 1e18; 8 const int maxn = 6e4; 9 const int maxm = 1e6; 10 int cnt, h[maxn]; 11 struct edge { 12 int to, pre, cap, cost; 13 } e[maxm << 1]; 14 void init() { 15 cnt = 0; 16 memset(h, -1, sizeof(h)); 17 } 18 void add(int from, int to, int cap, int cost) { 19 e[cnt].pre = h[from]; 20 e[cnt].to = to; 21 e[cnt].cap = cap; 22 e[cnt].cost = cost; 23 h[from] = cnt; 24 cnt++; 25 } 26 ll dist[maxn]; 27 int vis[maxn]; 28 int pv[maxn], pe[maxn]; 29 ll min_cost_flow(int s, int t, int f) { 30 ll ret = 0; 31 while (f > 0) { 32 memset(vis, 0, sizeof(vis)); 33 fill(dist, dist + maxn, inf); 34 dist[s] = 0; 35 queue<int> q; 36 q.push(s); 37 while (!q.empty()) { 38 int v = q.front(); 39 q.pop(); 40 vis[v] = 0; 41 for (int i = h[v]; i >= 0; i = e[i].pre) { 42 int to = e[i].to, cap = e[i].cap; 43 ll cost = e[i].cost; 44 if (cap > 0 && dist[to] > dist[v] + cost) { 45 pv[to] = v, pe[to] = i; 46 dist[to] = dist[v] + cost; 47 if (!vis[to]) q.push(to); 48 vis[to] = 1; 49 } 50 } 51 } 52 53 if (dist[t] > 0) return ret; // modify here 54 55 int d = f; 56 for (int v = t; v != s; v = pv[v]) 57 d = min(d, e[pe[v]].cap); 58 f -= d; 59 ret += d * dist[t]; 60 for (int v = t; v != s; v = pv[v]) { 61 e[pe[v]].cap -= d; 62 e[pe[v] ^ 1].cap += d; 63 } 64 } 65 return ret; 66 } 67 68 int main() { 69 int r, c; 70 scanf("%d %d", &r, &c); 71 ll ans = 0; 72 for(int i = 1; i <= r; ++i) { 73 for(int j = 1; j <= c; ++j) { 74 scanf("%d", &a[i][j]); 75 if(a[i][j]) ans += a[i][j] - 1; 76 } 77 } 78 init(); 79 int S = r + c + 1, T = S + 1, id = T; 80 for(int i = 1; i <= r; ++i) { 81 for(int j = 1; j <= c; ++j) { 82 rmax[i] = max(rmax[i], a[i][j]); 83 } 84 if(rmax[i]) { 85 ans -= rmax[i] - 1; 86 add(S, i, 1, 1 - rmax[i]); 87 add(i, S, 0, rmax[i] - 1); 88 } 89 } 90 for(int j = 1; j <= c; ++j) { 91 for(int i = 1; i <= r; ++i) { 92 cmax[j] = max(cmax[j], a[i][j]); 93 } 94 if(cmax[j]) { 95 ans -= cmax[j] - 1; 96 add(r + j, T, 1, 0); 97 add(T, r + j, 0, 0); 98 } 99 } 100 for(int i = 1; i <= r; ++i) { 101 for(int j = 1; j <= c; ++j) { 102 if(a[i][j] && rmax[i] == cmax[j]) { 103 ++id; 104 add(i, id, 1, 0), add(id, i, 0, 0); 105 add(id, r + j, 1, 0), add(r + j, id, 0, 0); 106 } 107 } 108 } 109 printf("%lld ", ans - min_cost_flow(S, T, 1e9)); 110 return 0; 111 }
从上到下扫描线每次匹配最右的不然肯定重叠了，最后检查一遍，注意要处理一下同一行内的重叠
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int maxn = 2e5 + 10; 4 int r[maxn], c[maxn], ans[maxn]; 5 vector<int> bx, by, row[maxn]; 6 map<int, int> col; 7 map<int, int> :: iterator it; 8 9 typedef pair<int, int> pii; 10 vector<pii> ins[maxn], del[maxn]; 11 set<pii> seg, rseg; 12 set<pii> :: iterator itl, itr; 13 inline pii re(pii x) { 14 return pii(-x.second, -x.first); 15 } 16 17 int main() { 18 int n; 19 scanf("%d", &n); 20 for(int i = 1; i <= n + n; ++i) { 21 scanf("%d %d", r + i, c + i); 22 bx.push_back(r[i]); 23 by.push_back(c[i]); 24 } 25 sort(bx.begin(), bx.end()); 26 bx.erase(unique(bx.begin(), bx.end()), bx.end()); 27 sort(by.begin(), by.end()); 28 by.erase(unique(by.begin(), by.end()), by.end()); 29 for(int i = 1; i <= n + n; ++i) { 30 r[i] = lower_bound(bx.begin(), bx.end(), r[i]) - bx.begin() + 1; 31 c[i] = lower_bound(by.begin(), by.end(), c[i]) - by.begin() + 1; 32 row[r[i]].push_back(i); 33 } 34 int ok = 1; 35 for(int i = 1; i <= bx.size(); ++i) { 36 for(int j = 0; j < row[i].size(); ++j) { 37 int x = row[i][j]; 38 if(x <= n) { 39 if(col.find(c[x]) == col.end()) col[c[x]] = x; 40 else ok = 0; 41 } 42 else { 43 it = col.upper_bound(c[x]); 44 if(col.empty() || it == col.begin()) ok = 0; 45 else { 46 --it; 47 int y = (*it).second; 48 ans[y] = x; 49 ins[r[y]].emplace_back(pii(c[y], c[x])); 50 del[r[x]].emplace_back(pii(c[y], c[x])); 51 col.erase(it); 52 } 53 } 54 } 55 } 56 if(ok) { 57 for(int i = 1; i <= bx.size(); ++i) { 58 sort(ins[i].begin(), ins[i].end()); 59 for(int j = 0; j < ins[i].size(); ++j) { 60 if(j && ins[i][j].first <= ins[i][j - 1].second) ok = 0; 61 seg.insert(ins[i][j]), rseg.insert(re(ins[i][j])); 62 itr = seg.upper_bound(ins[i][j]); 63 if(itr != seg.end()) { 64 pii tmp = *itr; 65 if(tmp.first <= ins[i][j].second) ok = 0; 66 } 67 itl = rseg.upper_bound(re(ins[i][j])); 68 if(itl != seg.end()) { 69 pii tmp = *itl; 70 if(tmp.first <= re(ins[i][j]).second) ok = 0; 71 } 72 } 73 sort(del[i].begin(), del[i].end()); 74 for(int j = 0; j < del[i].size(); ++j) { 75 if(j && del[i][j].first <= del[i][j - 1].second) ok = 0; 76 seg.erase(del[i][j]), rseg.erase(re(del[i][j])); 77 } 78 } 79 } 80 if(!ok) puts("syntax error"); 81 else for(int i = 1; i <= n; ++i) printf("%d ", ans[i] - n); 82 return 0; 83 }
首先只保留递减的点，然后决策单调性，就可以分治了
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int maxn = 1e6 + 10; 5 int p[maxn], d[maxn], id[maxn]; 6 7 bool cmp(int i, int j) { 8 return d[i] < d[j]; 9 } 10 11 ll ans; 12 vector<int> L, R; 13 inline ll cal(int i, int j) { 14 return 1ll * (d[R[j]] - d[L[i]]) * (p[R[j]] - p[L[i]]); 15 } 16 void solve(int l1, int r1, int l2, int r2) { 17 if(l2 > r2) return; 18 int m2 = (l2 + r2) / 2; 19 ll mx = -1e18, pos = -1; 20 for(int i = l1; i <= r1; ++i) { 21 if(d[L[i]] >= d[R[m2]]) continue; 22 ll tmp = cal(i, m2); 23 if(tmp > mx) mx = tmp, pos = i; 24 } 25 ans = max(mx, ans); 26 if(l2 != r2) { 27 if(pos == -1) solve(l1, r1, m2 + 1, r2); 28 else solve(l1, pos, l2, m2 - 1), solve(pos, r1, m2 + 1, r2); 29 } 30 } 31 32 int main() { 33 int m, n; 34 scanf("%d %d", &m, &n); 35 for(int i = 1; i <= m + n; ++i) { 36 scanf("%d %d", p + i, d + i); 37 id[i] = i; 38 } 39 sort(id + 1, id + 1 + m, cmp); 40 for(int i = 1; i <= m; ++i) { 41 if(!L.empty() && p[L.back()] <= p[id[i]]) continue; 42 L.push_back(id[i]); 43 } 44 sort(id + m + 1, id + m + 1 + n, cmp); 45 for(int i = n + m; i >= m + 1; --i) { 46 if(!R.empty() && p[R.back()] >= p[id[i]]) continue; 47 R.push_back(id[i]); 48 } 49 reverse(R.begin(), R.end()); 50 solve(0, L.size() - 1, 0, R.size() - 1); 51 printf("%lld ", ans); 52 return 0; 53 }
枚举两个顶点，判断是否是在多边形内且中间没有别的顶点，是的话往两边延伸，比较麻烦的是碰到顶点的情况，更麻烦的是顶点的一边与延伸方向重合的情况……
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 typedef long double db; 5 const db eps=1e-6; 6 const db inf=1e18; 7 int sign(db k){ 8 if (k>eps) return 1; else if (k<-eps) return -1; return 0; 9 } 10 int cmp(db k1,db k2){return sign(k1-k2);} 11 int inmid(db k1,db k2,db k3){return sign(k1-k3)*sign(k2-k3)<=0;}// k3 在 [k1,k2] 内 12 struct point{ 13 db x,y; 14 point operator + (const point &k1) const{return (point){k1.x+x,k1.y+y};} 15 point operator - (const point &k1) const{return (point){x-k1.x,y-k1.y};} 16 point operator * (db k1) const{return (point){x*k1,y*k1};} 17 point operator / (db k1) const{return (point){x/k1,y/k1};} 18 int operator == (const point &k1) const{return cmp(x,k1.x)==0&&cmp(y,k1.y)==0;} 19 bool operator < (const point k1) const{ 20 int a=cmp(x,k1.x); 21 if (a==-1) return 1; else if (a==1) return 0; else return cmp(y,k1.y)==-1; 22 } 23 db abs(){return sqrt(x*x+y*y);} 24 }; 25 int inmid(point k1,point k2,point k3){return inmid(k1.x,k2.x,k3.x)&&inmid(k1.y,k2.y,k3.y);} 26 db cross(point k1,point k2){return k1.x*k2.y-k1.y*k2.x;} 27 db dot(point k1,point k2){return k1.x*k2.x+k1.y*k2.y;} 28 int checkLL(point k1,point k2,point k3,point k4){// 求直线 (L) 线段 (S)k1,k2 和 k3,k4 的交点 29 return cmp(cross(k3-k1,k4-k1),cross(k3-k2,k4-k2))!=0; 30 } 31 point getLL(point k1,point k2,point k3,point k4){ 32 db w1=cross(k1-k3,k4-k3),w2=cross(k4-k3,k2-k3); return (k1*w2+k2*w1)/(w1+w2); 33 } 34 int intersect(db l1,db r1,db l2,db r2){ 35 if (l1>r1) swap(l1,r1); if (l2>r2) swap(l2,r2); return cmp(r1,l2)!=-1&&cmp(r2,l1)!=-1; 36 } 37 int checkSS(point k1,point k2,point k3,point k4){ 38 return intersect(k1.x,k2.x,k3.x,k4.x)&&intersect(k1.y,k2.y,k3.y,k4.y)&& 39 sign(cross(k3-k1,k4-k1))*sign(cross(k3-k2,k4-k2))<=0&& 40 sign(cross(k1-k3,k2-k3))*sign(cross(k1-k4,k2-k4))<=0; 41 } 42 int onS(point k1,point k2,point q){return inmid(k1,k2,q)&&sign(cross(k1-q,k2-k1))==0;} 43 int contain(vector<point>A,point q){ // 2 内部 1 边界 0 外部 44 int pd=0; A.push_back(A[0]); 45 for (int i=1;i<A.size();i++){ 46 point u=A[i-1],v=A[i]; 47 if (onS(u,v,q)) return 1; if (cmp(u.y,v.y)>0) swap(u,v); 48 if (cmp(u.y,q.y)>=0||cmp(v.y,q.y)<0) continue; 49 if (sign(cross(u-v,q-v))<0) pd^=1; 50 } 51 return pd<<1; 52 } 53 54 int main() { 55 int n; 56 scanf("%d", &n); 57 vector<point> p(n); 58 for(int i = 0; i < n; ++i) cin >> p[i].x >> p[i].y; 59 double ans = 0; 60 for(int i = 0; i < n; ++i) { 61 for(int j = i + 1; j < n; ++j) { 62 int ok = 1; 63 for(int k = 0; k < n; ++k) { 64 if(i == k && j == (k + 1) % n || j == k && i == (k + 1) % n) continue; 65 if(checkSS(p[i], p[j], p[k], p[(k + 1) % n])) { 66 if(onS(p[i], p[j], p[k]) && onS(p[i], p[j], p[(k + 1) % n])) {ok = 0; break;} 67 point x = getLL(p[i], p[j], p[k], p[(k + 1) % n]); 68 if(x == p[i] || x == p[j]) continue; 69 ok = 0; break; 70 } 71 } 72 point mid = (p[i] + p[j]) / 2; 73 if(!contain(p, mid)) ok = 0; 74 if(!ok) continue; 75 vector<point> q; 76 for(int k = 0; k < n; ++k) { 77 if(checkLL(p[i], p[j], p[k], p[(k + 1) % n])) { 78 point x = getLL(p[i], p[j], p[k], p[(k + 1) % n]); 79 if(onS(p[k], p[(k + 1) % n], x)) { 80 if(x == p[k]) { 81 point& nxt = p[(k + 1) % n]; 82 point& pre = p[(k + n - 1) % n]; 83 if(sign(cross(p[j] - p[i], pre - p[i])) == 0) { 84 if(sign(dot(p[k] - pre, p[j] - p[i])) > 0 && sign(cross(p[j] - p[i], nxt - p[i])) > 0) q.push_back(x); 85 if(sign(dot(p[k] - pre, p[i] - p[j])) > 0 && sign(cross(p[i] - p[j], nxt - p[j])) > 0) q.push_back(x); 86 } 87 else { 88 if(sign(cross(p[j] - p[i], pre - p[i])) < 0 && sign(cross(p[j] - p[i], nxt - p[i])) > 0) q.push_back(x); 89 if(sign(cross(p[i] - p[j], pre - p[j])) < 0 && sign(cross(p[i] - p[j], nxt - p[j])) > 0) q.push_back(x); 90 } 91 } 92 else if(x == p[(k + 1) % n]) { 93 point& nxt = p[(k + 2) % n]; 94 point& cur = p[(k + 1) % n]; 95 if(sign(cross(p[j] - p[i], nxt - p[i])) == 0) { 96 if(sign(dot(nxt - cur, p[j] - p[i])) < 0 && sign(cross(p[j] - p[i], p[k] - p[i])) < 0) q.push_back(x); 97 if(sign(dot(nxt - cur, p[i] - p[j])) < 0 && sign(cross(p[i] - p[j], p[k] - p[j])) < 0) q.push_back(x); 98 } 99 else { 100 if(sign(cross(p[j] - p[i], nxt - p[i])) > 0 && sign(cross(p[j] - p[i], p[k] - p[i])) < 0) q.push_back(x); 101 if(sign(cross(p[i] - p[j], nxt - p[j])) > 0 && sign(cross(p[i] - p[j], p[k] - p[j])) < 0) q.push_back(x); 102 } 103 } 104 else q.push_back(x); 105 } 106 } 107 } 108 db di = inf, dj = inf; 109 for(int k = 0; k < q.size(); ++k) { 110 di = min(di, (p[i] - q[k]).abs()); 111 dj = min(dj, (p[j] - q[k]).abs()); 112 } 113 ans = max(ans, double((p[i] - p[j]).abs() + di + dj)); 114 } 115 } 116 printf("%.8f ", ans); 117 return 0; 118 }
不会猜结论
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int maxn = 1e5 + 10; 4 vector<int> v[11]; 5 string str[11]; 6 char ch[maxn]; 7 int id[11]; 8 9 int nxt[maxn]; 10 inline void getnxt(int x) { 11 nxt[0] = nxt[1] = 0; 12 int l = str[x].length(); 13 for (int i = 1; i < l; i++) { 14 int j = nxt[i]; 15 while (j && str[x][i] != str[x][j]) j = nxt[j]; 16 nxt[i + 1] = str[x][i] == str[x][j] ? j + 1 : 0; 17 } 18 } 19 20 bool cmp(int i, int j) { 21 int sz = min(v[i].size(), v[j].size()); 22 for(int k = 0; k < sz; ++k) if(v[i][k] != v[j][k]) return v[i][k] > v[j][k]; 23 if(v[i].size() != v[j].size()) return v[i].size() < v[j].size(); 24 return i < j; 25 } 26 27 int main() { 28 int n, s; 29 scanf("%d %d", &n, &s); 30 for(int i = 1; i <= s; ++i) { 31 scanf("%s", ch); 32 str[i] = string(ch); 33 getnxt(i); 34 int l = str[i].length(), p = l; 35 while(p) { 36 p = nxt[p]; 37 if(l + l - p <= n) v[i].push_back(l + l - p); 38 } 39 id[i] = i; 40 } 41 sort(id + 1, id + 1 + s, cmp); 42 for(int i = 1; i <= s; ++i) printf("%s ", str[id[i]].c_str()); 43 return 0; 44 }
不会观察
1 #include <bits/stdc++.h> 2 using namespace std; 3 int f[333][333], g[333][333]; 4 5 #define F(i, j) ((i < u || j < l) ? 0 : f[i][j]) 6 #define G(i, j) ((i < u || j < l) ? 0 : g[i][j]) 7 inline int check_rows(int u, int d, int l, int r) { 8 for(int i = u; i <= d + 1; ++i) { 9 for(int j = l; j <= r + 1; ++j) { 10 g[i][j] = F(i - 1, j - 1); 11 for(int di = -1; di <= 1; ++di) { 12 for(int dj = -1; dj <= 1; ++dj) { 13 if(di == 1 && dj == 1) continue; 14 g[i][j] ^= G(i - 1 + di, j - 1 + dj); 15 } 16 } 17 } 18 if(g[i][r] || g[i][r + 1]) return i - 1; 19 } 20 return -1; 21 } 22 inline int check_cols(int u, int d, int l, int r) { 23 for(int j = l; j <= r + 1; ++j) { 24 for(int i = u; i <= d + 1; ++i) { 25 g[i][j] = F(i - 1, j - 1); 26 for(int di = -1; di <= 1; ++di) { 27 for(int dj = -1; dj <= 1; ++dj) { 28 if(di == 1 && dj == 1) continue; 29 g[i][j] ^= G(i - 1 + di, j - 1 + dj); 30 } 31 } 32 } 33 if(g[d][j] || g[d + 1][j]) return j - 1; 34 } 35 return -1; 36 } 37 inline void shrink(int& u, int& d, int& l, int& r) { 38 for(int i = u; i <= d; ++i) { 39 int ok = 1; 40 for(int j = l; j <= r; ++j) 41 if(f[i][j]) ok = 0; 42 if(ok) u++; 43 else break; 44 } 45 for(int i = d; i >= u; --i) { 46 int ok = 1; 47 for(int j = l; j <= r; ++j) 48 if(f[i][j]) ok = 0; 49 if(ok) d--; 50 else break; 51 } 52 for(int j = l; j <= r; ++j) { 53 int ok = 1; 54 for(int i = u; i <= d; ++i) 55 if(f[i][j]) ok = 0; 56 if(ok) l++; 57 else break; 58 } 59 for(int j = r; j >= l; --j) { 60 int ok = 1; 61 for(int i = u; i <= d; ++i) 62 if(f[i][j]) ok = 0; 63 if(ok) r--; 64 else break; 65 } 66 } 67 68 int main() { 69 int n, m; 70 scanf("%d %d", &m, &n); 71 for(int i = 1; i <= n; ++i) { 72 char s[333]; 73 scanf("%s", s + 1); 74 for(int j = 1; j <= m; ++j) f[i][j] = s[j] == '#'; 75 } 76 int u = 1, d = n, l = 1, r = m; 77 shrink(u, d, l, r); 78 while(d > u && r > l) { 79 int row = check_rows(u, d, l, r); 80 if(row == -1) memcpy(f, g, sizeof(f)); 81 else { 82 int col = check_cols(u, d, l, r); 83 f[row][col] ^= 1; 84 int t = check_rows(u, d, l, r); 85 if(t == -1) memcpy(f, g, sizeof(f)); 86 else {f[row][col] ^= 1; break;} 87 } 88 shrink(u, d, l, r); 89 } 90 for(int i = u; i <= d; ++i) { 91 for(int j = l; j <= r; ++j) putchar(f[i][j] ? '#' : '.'); 92 puts(""); 93 } 94 return 0; 95 }