Hunan multi-school competition (2015.4.6) CSU 1561 ~ 1569 question solution, 2015.4.6csu

Hunan multi-school competition (2015.4.6) CSU 1561 ~ 1569 question solution, 2015.4.6csu

A: Click to open the link.

CSU 1561 (More) Multiplication

Fill in the result of the product of two numbers at each position.

Note that the maximum size of this matrix is 1e8, so the array cannot be opened.

Simulation questions

#include <iostream>#include <cstdio>#include <algorithm>#include <string>#include <cmath>#include <cstring>#include <queue>#include <set>#include <map>#include <vector>using namespace std;typedef long long ll;const int N = 10005;template <class T>inline bool rd(T &ret) {char c; int sgn;if (c = getchar(), c == EOF) return 0;while (c != '-' && (c<'0' || c>'9')) c = getchar();sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c - '0');ret *= sgn;return 1;}template <class T>inline void pt(T x) {if (x <0) {putchar('-');x = -x;}if (x>9) pt(x / 10);putchar(x % 10 + '0');}int ans[N * 2];char A[N], B[N];char a[N * 4];int main() {while (~scanf("%s %s", A, B)) {int m = strlen(A), n = strlen(B), x = 0, y = 0;for (int i = 0; i < m; i++) {x = x * 10 + (A[i] - '0');}for (int i = 0; i < n; i++) {y = y * 10 + (B[i] - '0');}if (x == 0 && y == 0) break;int num = x*y, size = 0;while (num > 0) {ans[size++] = num % 10;num /= 10;}while (size < n + m) {ans[size++] = -1;}n = n * 4 + 5;m = m * 4 + 5;bool f = false;for (int ii = 0; ii < n; ii++) {a[m] = '\0';for (int jj = 0; jj < m; jj++) a[jj] = ' ';if (ii == 0 || ii == n - 1) {a[0] = a[m - 1] = '+';for (int jj = 1; jj < m - 1; jj++) {a[jj] = '-';}}else {a[0] = a[m - 1] = '|';if (ii == 1) {for (int j = 4; j < m - 1; j += 4) {a[j] = A[j / 4 - 1];}}else if (ii == n - 2) {for (int pos = m - 6, id = 0; pos > 0; pos -= 4, id++) {a[pos] = ans[id] + '0';a[pos - 2] = '/';}if (!f) a[1] = ' ';}else {if ((ii - 2) % 4 == 0) {for (int j = 2; j < m - 2; j++) {if ((j - 2) % 4 == 0) a[j] = '+';else a[j] = '-';}}else {for (int j = 2; j < m - 2; j += 4) {a[j] = '|';}if ((ii + 1) % 4 == 0) {if (f) a[1] = '/';for (int j = 4; j < m - 1; j += 4) {a[j + 1] = '/';a[j - 1] = (A[j / 4 - 1] - '0') * (B[(ii + 1) / 4 - 1] - '0') / 10 + '0';}}if (ii % 4 == 0) {for (int j = 4; j < m - 1; j += 4) {a[j] = '/';}a[m - 2] = B[ii / 4 - 1];}if ((ii - 1) % 4 == 0) {if (ans[size - 1] != -1) {a[1] = ans[size - 1] + '0';f = true;}size--;for (int j = 4; j < m - 1; j += 4) {a[j - 1] = '/';a[j + 1] = (A[j / 4 - 1] - '0') * (B[(ii - 1) / 4 - 1] - '0') % 10 + '0';}}}}}printf("%s\n", a);}}return 0;}

B: Click the open link.
CSU 1562 Fun House

Question:

Given a matrix, there are two kinds of mirrors/\ inside the Matrix. If a ray is incident from the edge, the light will surely hit the edge and mark the edge.

Then output this matrix.

#include <iostream>#include <cstdio>#include <algorithm>#include <string>#include <cmath>#include <cstring>#include <queue>#include <set>#include <map>#include <vector>using namespace std;const int MAX_N = 27;int n, m;int ex, ey;char maz[27][27];bool vis[MAX_N][MAX_N];bool found;bool safe(int x, int y) {return x >= 0 && x < m && y >= 0 && y < n;}void dfs(int x, int y, int dir) {if (found) return ;if (dir == 1) ++x;else if (dir == 2) ++y;else if (dir == 3) --x;else --y;if (safe(x, y)) {if (maz[x][y] == 'x') {ex = x, ey = y;found = true;return ;}if (maz[x][y] == '.') dfs(x, y, dir);else if (maz[x][y] == '/') {if (dir == 1) {dfs(x, y, 4);} else if (dir == 2) {dfs(x, y, 3);} else if (dir == 3) {dfs(x, y, 2);} else {dfs(x, y, 1);}} else {if (dir == 1) {dfs(x, y, 2);} else if (dir == 2) {dfs(x, y, 1);} else if (dir == 3) {dfs(x, y, 4);} else {dfs(x, y, 3);}}}}void Clear() {ex = 0, ey = 0;found = false;memset(maz, 0, sizeof maz);}int main() {int cas = 0;while (2 == scanf("%d%d", &n, &m)) {if (0 == n && 0 == m) break;int sx = 0, sy = 0;Clear();for (int i = 0; i < m; ++i) {scanf("%s", maz[i]);for (int j = 0; j < n; ++j) {if (maz[i][j] == '*') {sx = i, sy = j;maz[i][j] = '.';}}}if (sx == 0)dfs(sx, sy, 1);else if (sx == m-1)dfs(sx, sy, 3);else if (sy == 0)dfs(sx, sy, 2);elsedfs(sx, sy, 4);printf("HOUSE %d\n", ++cas);for (int i = 0; i < m; ++i) {for (int j = 0; j < n; ++j) {if (i == sx && j == sy) {putchar('*');} else if (i == ex && j == ey) {putchar('&');} else putchar(maz[i][j]);}puts("");}}    return 0;}

C: Click the open link.
Question:

Given a string, ask what is the smallest k in the Lexicographic Order.

From high to low, the maximum number of letters filled in each time <= k.

#include <iostream>#include <cstdio>#include <algorithm>#include <string>#include <cmath>#include <cstring>#include <queue>#include <set>#include <map>#include <vector>using namespace std;typedef long long ll;const int N = 20;char s[N], t[N];ll k;ll num[27];ll J[30];int len;ll way(){ll sum = 0;for(int i = 0; i < 26; i++)sum += num[i];ll ans = J[sum];for(int i= 0; i < 26; i++)ans /= J[num[i]];return ans;}void dfs(int now , ll pre){if(now == len)return ;char c = 'A';for(int i = 0; i < 26; i++){if(num[i]==0)continue;c = 'A'+i;num[i]--;ll hehe = way();num[i]++;if(pre + hehe >= k){break;}pre += hehe;}num[c-'A']--;t[now] = c;dfs(now+1, pre);}int main() {J[0] = 1;for(ll i = 1; i < 30; i++)J[i] = J[i-1]*i;while (~scanf("%s", s)) {cin>>k;if(s[0] == '#' && k==0)break;memset(num, 0, sizeof num);len = strlen(s);memset(t, 0, sizeof t);for(int i = 0; s[i]; i++)num[s[i]-'A'] ++;dfs(0, 0);puts(t);}return 0;}

E: Click the open link.

CSU 1565 Word Cloud

A simulation question.

The width of the matrix is P and N letters are given.

Each letter and its height are given below.

Use the figure in the question to find the letter width.

Fill letters in the matrix from top to bottom, from left to right, and ask what the height of the matrix is after filling.

Note that the distance between letters is 10px.

#include <iostream>#include <cstdio>#include <algorithm>#include <string>#include <cmath>#include <cstring>#include <queue>#include <set>#include <map>#include <vector>using namespace std;const int MAX_N = 107;int w, n;char str[100007];int c[MAX_N], len[MAX_N];int H[MAX_N], W[MAX_N];int main() {int cas = 0;while (2 == scanf("%d%d", &w, &n)) {if (0 == w && 0 == n) break;int Cmax = 0;for (int i = 0; i < n; ++i) {scanf("%s%d", str, &c[i]);Cmax = max(Cmax, c[i]);len[i] = strlen(str);}for (int i = 0; i < n; ++i) {H[i] = 8 + ceil(40 * (c[i] - 4) * 1. / (Cmax - 4));W[i] = ceil(9. * len[i] * H[i] / 16);}long long ans = 0, sum = 0;bool row = false;int maxH = 0;w += 10;for (int i = 0; i < n; ++i) {if (row) {row = false;ans += maxH;sum = 0;maxH = 0;}if (sum + W[i] + 10 <= w) {maxH = max(maxH, H[i]);sum += W[i] + 10;} else row = true, --i;}if (sum != 0) ans += maxH;printf("CLOUD %d: %lld\n", ++cas, ans);}    return 0;}

F: Click the open link.
CSU 1566 The Maze Makers

Given a labyrinth graph, each grid has 16 forms, represented by a hexadecimal number.

The edge has only two gaps.

Q:

If two gaps are not connected, the output is "No solution"

If else haS a vertex that is not reachable, the output "Unreachable cell"

Else if there are different paths between the two gaps, the output is "Multiple path"

Else "Maze OK"

#include <iostream>#include <cstdio>#include <algorithm>#include <string>#include <cmath>#include <cstring>#include <queue>#include <set>#include <map>#include <vector>using namespace std;typedef long long ll;typedef pair<int,int> pii;const int N = 55;int n, m;vector<int>G[N*N], path;vector<pii>D;int has(int x, int y){return x*m+y;}void add(int x, int y, int i, int j){G[has(x,y)].push_back(has(i,j));}int getx(int val){return val/m;}int gety(int val){return val%m;}bool inmap(int x, int y){return 0<=x && x < n && 0<=y&&y<m;}char s[N];int mp[N][N];int step[4][2] = {0,-1, 1,0, 0,1, -1,0};bool vis[N][N];void bfs(int x, int y){memset(vis, 0, sizeof vis);queue<int>q;q.push(has(x,y));vis[x][y] = true;while(!q.empty()){int u = q.front(); q.pop();for(int i = 0; i < G[u].size(); i++){int v = G[u][i];x = getx(v); y = gety(v);if(vis[x][y])continue;vis[x][y] = true;q.push(v);}}}int solve(){if(false == vis[D[1].first][D[1].second])return 0;for(int i = 0; i < n; i++)for(int j = 0; j < m; j++)if(false == vis[i][j])return 1;int siz = 0;for(int i = 0; i < n; i++)for(int j = 0; j < m; j++)siz += G[has(i,j)].size();if((siz>>1)==n*m-1)return 3;return 2;}int main() {while (~scanf("%d %d", &n, &m), n+m) {D.clear();for(int i = 0; i < n; i++){scanf("%s", s);for(int j = 0; j < m; j++){G[has(i,j)].clear();if('0' <= s[j] && s[j]<='9')mp[i][j] = s[j]-'0';else mp[i][j] = s[j]-'A'+10;}}for(int i = 0; i < n; i++)for(int j = 0; j < m; j++)for(int k = 0; k < 4; k++){int x = i+step[k][0], y = j+step[k][1];if(!inmap(x,y)){if((mp[i][j]&(1<<k))==0)D.push_back(pii(i,j));continue;}if((mp[i][j]&(1<<k))==0)add(i,j, x,y);}sort(D.begin(), D.end());D.erase(unique(D.begin(), D.end()), D.end());if((int)D.size()==1) D.push_back(D[0]);bfs(D[0].first, D[0].second);int ans = solve();if(ans == 0)puts("NO SOLUTION");else if(ans == 1) puts("UNREACHABLE CELL");else if(ans == 2)puts("MULTIPLE PATHS");else puts("MAZE OK");}return 0;}/*1 82AAAAAAE*/

G: Click to open the link.
CSU 1567 Reverse Rot

#include <iostream>#include <cstdio>#include <algorithm>#include <string>#include <cmath>#include <cstring>#include <queue>#include <set>#include <map>#include <vector>using namespace std;const int MAX_N = 57;const char table[30] = "ABCDEFGHIJKLMNOPQRSTUVWXYZ_.";int n;char str[MAX_N];int ID(char ch) {for (int i = 0; table[i]; ++i) if (table[i] == ch) return i;}int main() {   while (1 == scanf("%d", &n)) {if (0 == n) break;scanf("%s", str);for (int i = 0; str[i]; ++i) {str[i] = table[(ID(str[i]) + n) % 28];}int len = strlen(str);for (int i = len - 1; i >= 0; --i) {putchar(str[i]);}puts("");   }    return 0;}

H: Click the open link.

CSU 1568 Shrine Maintenance

Enter m n, D, and D.

Divide a circle with a radius of 1000 into N equal points, use all the valid points in the N points on the circle to draw exactly M class triangles (I want to say that there are exactly two sides of a class triangle connected to the Circle Center, even if only one vertex is connected in the second graph)

Effective point: we give the vertex number in the circle (1 ~ N), so the multiple of any number in the D number is the effective point.

Q:

Minimize the maximum perimeter of M class triangles drawn. Output the perimeter and keep a decimal number.

Ideas:

The second answer is feasible.

We connect adjacent valid points to each other.

W is equivalent to disconnecting W valid edges.

If W = 0, all edges are connected for a long time, so it is a ring.

The first breakpoint of the enumeration ring is used, but t

However, we find that the interval between the preceding factors is relatively large, and the interval is the same after the lcm is reached. In this way, disconnecting any side is equivalent. That is to say, enumerate the preceding factor.

#include <iostream>#include <cstdio>#include <algorithm>#include <string>#include <cmath>#include <cstring>#include <queue>#include <set>#include <map>#include <vector>using namespace std;const int MAX_N = 8607;int W, N, D;int cnt;bool vis[MAX_N];int id[MAX_N];void Clear() {cnt = 0;memset(vis, false, sizeof vis);}double get(int i, int j) {return 2000. * sin(acos(-1.) * abs(id[j] - id[i]) / N);}bool check(double ans, int j) {int w = W;double sum = 0;for (int i = (j+2)%cnt; ; i = (i+1)%cnt) {double now = get((i - 1+cnt)%cnt, i);if (sum + now <= ans) {sum += now;}else {sum = 0;--w;if (w < 0) return false;}if(i==j)break;}return true;}int main() {while (1 == scanf("%d", &W)) {if (0 == W) break;Clear();scanf("%d%d", &N, &D);for (int i = 0; i < D; ++i) {int v;scanf("%d", &v);for (int j = v; j <= N; j += v)vis[j] = true;}for (int i = 1; i <= N; ++i) if (vis[i])id[cnt++] = i;if(cnt <= W) {puts("2000.0");continue;}if(W == 0){puts("0.0"); continue;}W--;double ans = 1e8;for(int j = 0; j < min(100,cnt); j++){double l = 0, r = ans;for (int i = 0; i <30; ++i) {double mid = (l + r) / 2;if (check(mid, j)) r = mid;else l = mid;}ans = min(ans, r);}printf("%.1f\n", ans + 2000);}    return 0;}

I: Click to open the link.

CSU 1569 Wet Tiles

Given n * m matrix moment t L start point W walls

The coordinates of L points are given below.

Each 4 numbers below give the start and end points of each wall (the wall is either parallel to the axis or 45 ° tilt)

Ask the number of vertices that the entire graph has traversed after each point bfs goes out under t.

Simple bfs

#include <cstdio>#include <cstring>#include <queue>#include <algorithm>using namespace std;typedef pair<int, int> pii;const int N = 1005;char mp[N][N];int vis[N][N];int n, m, T;queue<pii> que;const int dx[] = {0, 0, -1, 1};const int dy[] = {-1, 1, 0, 0};int bfs() {int ans = 0;while(que.size() > 0) {pii tmp = que.front(); que.pop();int x = tmp.first, y = tmp.second;ans ++;if(vis[x][y] == T) continue;for(int i = 0; i < 4; i ++) {int nx = x + dx[i];int ny = y + dy[i];if(nx >= 0 && nx < n && ny >= 0 && ny < m && vis[nx][ny] == 0) {vis[nx][ny] = vis[x][y] + 1;que.push(make_pair(nx, ny));}}}return ans;}int main() {while(~scanf("%d", &m)) {if(m == -1) break;memset(vis, 0, sizeof vis);int L, W;scanf("%d%d%d%d", &n, &T, &L, &W);for(int i = 0, x, y; i < L; i ++) {scanf("%d%d", &x, &y);swap(x, y);  y--, x = n - x;que.push(make_pair(x, y));vis[x][y] = 1;}for(int i = 0, x1, x2, y1, y2; i < W; i ++) {scanf("%d%d%d%d", &x1, &y1, &x2, &y2);swap(x1, y1); swap(x2, y2);  y1--, y2--;x1 = n - x1;x2 = n - x2;if(x1 == x2) {if(y1 < y2) {while(y1 <= y2) {vis[x1][y1++] = -1;}} else {while(y1 >= y2) {vis[x1][y1--] = -1;}}} else if(y1 == y2) {if(x1 < x2) {while(x1 <= x2) {vis[x1++][y1] = -1;}} else {while(x1 >= x2) {vis[x1--][y1] = -1;}}} else {if(x1 < x2) {if(y1 < y2) {while(x1 <= x2) {vis[x1++][y1++] = -1;}} else {while(x1 <= x2) {vis[x1++][y1--] = -1;}}} else {if(y1 < y2) {while(x1 >= x2) {vis[x1--][y1++] = -1;}} else {while(x1 >= x2) {vis[x1--][y1--] = -1;}}}}}printf("%d\n", bfs());}return 0;}