题目链接
A:水题,把对应位置和标号一一匹配上,然后每次k多少对应就去坐多少即可
B:其实只要找周围4条,看能不能形成一个环,因为如果不成环,肯定有一个点会到不了,如果成环,那么中间任意点,就可以到达
C:找出左上角的点,然后以一边为长,枚举另一边的长度,然后dfs一遍,dfs的过程利用容斥可以加速
D:记录下gcd,每次多一个数字,就和前面的数字都取gcd,然后合并掉,下次利用这些已有的gcd去做,然后合并的过程要利用到gcd的递减性质,就可以直接从头往后找即可
代码:
A:
#include <cstdio> #include <cstring> #include <vector> #include <algorithm> #include <string> #include <iostream> using namespace std; int k; string ans[6]; int x[35], y[35]; int main() { ans[0] = "+------------------------+"; ans[1] = "|#.#.#.#.#.#.#.#.#.#.#.|D|)"; ans[2] = "|#.#.#.#.#.#.#.#.#.#.#.|.|"; ans[3] = "|#.......................|"; ans[4] = "|#.#.#.#.#.#.#.#.#.#.#.|.|)"; ans[5] = "+------------------------+"; x[0] = 1; y[0] = 1; x[1] = 2; y[1] = 1; x[2] = 3; y[2] = 1; x[3] = 4; y[3] = 1; int cnt = 3; for (int i = 4; i < 34; i++) { if ((i - 4) % 3 == 0) { x[i] = 1; y[i] = cnt; } if ((i - 4) % 3 == 1) { x[i] = 2; y[i] = cnt; } if ((i - 4) % 3 == 2) { x[i] = 4; y[i] = cnt; cnt += 2; } } int k; scanf("%d", &k); for (int i = 0; i < k; i++) ans[x[i]][y[i]] = 'O'; for (int i = 0; i < 6; i++) cout << ans[i] << endl; return 0; }
#include <cstdio> #include <cstring> #include <vector> using namespace std; int n, m; char a[25], b[25]; bool judge(char up, char down, char left, char right) { if (up == down) return false; if (left == right) return false; if (up == '<' && left == '^') return false; if (down == '<' && left == 'v') return false; if (up == '>' && right == '^') return false; if (down == '>' && right == 'v') return false; return true; } int main() { scanf("%d%d", &n, &m); scanf("%s%s", a, b); char up = a[0], down = a[n - 1]; char left = b[0], right = b[m - 1]; if (judge(up, down, left, right)) printf("YES\n"); else printf("NO\n"); return 0; }
#include <cstdio> #include <algorithm> #include <cstring> #include <queue> using namespace std; #define sum(x1,y1,x2,y2) (g[x2][y2] - g[x1 - 1][y2] - g[x2][y1 - 1] + g[x1 - 1][y1 - 1]) const int INF = 0x3f3f3f3f; const int N = 1005; int n, m, g[N][N], ans = INF; char str[N][N]; int dfs(int x,int y,int wx,int wy) { if(sum(x, y + 1, x + wx - 1, y + wy) == wx * wy) return wx + dfs(x, y + 1, wx, wy); if(sum(x + 1, y, x + wx, y + wy - 1) == wx * wy) return wy + dfs(x + 1, y, wx, wy); return 0; } int main() { scanf("%d%d",&n,&m); int flag = 0,px,py; for(int i = 1; i <= n; i++) scanf("%s",str[i] + 1); for(int i = 1; i <= n; i++) for(int j = 1; j <= m; j++) { if(str[i][j] == 'X') { if(!flag) {flag = 1; px = i; py = j;} g[i][j] = g[i - 1][j] + g[i][j - 1] - g[i - 1][j - 1] + 1; } else g[i][j] = g[i - 1][j] + g[i][j - 1] - g[i - 1][j - 1]; } int tmp,wx,wy; for(tmp = px; str[tmp][py] == 'X'; tmp++); wx = tmp - px; for(int i = py; str[px][i] == 'X'; i++) if(dfs(px, py, wx, i - py + 1) + wx * (i - py + 1) == g[n][m]) ans = min(ans, wx * (i - py + 1)); for(tmp = py; str[px][tmp] == 'X'; tmp++); wy = tmp - py; for(int i = px; str[i][py] == 'X'; i++) if(dfs(px, py, i - px + 1, wy) + (i - px + 1) * wy == g[n][m]) ans = min(ans, (i - px + 1) * wy); if (ans == INF) ans = -1; printf("%d\n", ans); return 0; }
#include <cstdio> #include <cstring> #include <algorithm> #include <map> using namespace std; typedef long long ll; typedef pair<int, ll> pii; const int N = 100005; int a[N]; pii save[N]; map<int, ll> ans; int n; int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); } void build() { int top = 0; for (int i = 1; i <= n; i++) { for (int j = 0; j < top; j++) save[j].first = gcd(save[j].first, a[i]); save[top++] = make_pair(a[i], 1LL); int sn = 1; for (int j = 1; j < top; j++) { if (save[sn - 1].first == save[j].first) save[sn - 1].second += save[j].second; else save[sn++] = save[j]; } top = sn; for (int j = 0; j < top; j++) { ans[save[j].first] += save[j].second; } } } int main() { scanf("%d", &n); for (int i = 1; i <= n; i++) scanf("%d", &a[i]); build(); int q; scanf("%d", &q); while (q--) { int x; scanf("%d", &x); printf("%lld\n", ans[x]); } return 0; }