總結——關於2017 10 23測試的分析總結

NOIP 2017 模拟

10 23

T1 :

题目:

——正解思路：

因为斐波那契数列增长很快，所以1e+9以内的数不超过50个，先预处理再O( 1 )判断即可。

——我的乱搞：

**没有乱搞，同正解。

tips:

珍惜这些弱智题，毕竟明天凯爷又要   hehehe   了。。。。

代码：

#include <iostream>#include <fstream>#include <map>int t, n, f[46] = {0, 1};std::map <int, bool> m;int main () {for (int i = 2; i < 45; i++) f[i] = f[i - 1] + f[i - 2];for (int i = 0; i < 45; i++) m[f[i]] = true;std::cin>>t;while (t--) {std::cin>>n;if (n == 0) {std::cout<<"Yes"<<std::endl; continue;}for (int i = 1; f[i] <= n and i < 45; i++) {if (n % f[i]) continue;if (m[n / f[i]]) {std::cout<<"Yes"<<std::endl; goto Con;}}std::cout<<"No"<<std::endl;Con : ;}return 0;}

T2 :

题目：

——题目正解：

LCA求中点。

——我的乱搞：

迪杰斯特拉暴搜。

tips:

中规中矩的暴力，不过在暴力之前貌似我又把LCA背拐了曳。。。。

#include <fstream>#define U(x, y) for (int i = x; i <= y; i++)#define Dfor (int i = 17; i >= 0; i--)#define N 100010int n, m, x, y, z, a, c;int t[N * 2], e[N * 2], d[N], s[N], r[N], f[N][18];void R (int &x) {x = 0;char c = 0;while (!isdigit(c)) c = getchar();while (isdigit(c)) x = x * 10 + c - 48, c = getchar();}void A (int x, int y) {t[++c] = y;e[c] = s[x];s[x] = c;}void T (int x, int F) {d[x] = d[F] + 1;f[x][0] = F;U(1, 17) f[x][i] = f[f[x][i - 1]][i - 1];for (int i = s[x]; i; i = e[i]) if (!d[t[i]]) {T (t[i], x);r[x] += r[t[i]];}}int L (int x, int y) {int E = d[x] - d[y];D if (E & (1 << i)) x = f[x][i];if (x == y) return x;D if (f[x][i] != f[y][i]) x = f[x][i], y = f[y][i];return f[x][0];}int main () {R(n);U(2, n) {R(x);R(y);A (x, y);A (y, x);r[i] = 1;}r[1] = 1;~d[0];T (1, 0);R(m);while (m--) {R(x);R(y);if (x == y) {printf ("%d\n", n);continue;}if (d[x] < d[y]) std::swap (x, y);z = L (x, y);c = d[x] + d[y] - d[z] * 2;if (c & 1) {puts("0");continue;}--c >>= 1;if (d[x] == d[y]) {D if (c & (1 << i)) x = f[x][i], y = f[y][i];a = r[1] - r[x] - r[y];} else {D if (c & (1 << i)) x = f[x][i];a = r[f[x][0]] - r[x];}printf ("%d\n", a);}return 0;}

T3 :

题目：

——正解思路：

其实各种方法都可以Ac，dp, dfs, 最大流神马的都可以。

——我的乱搞：

推出了dfs前一半思路，后面乱搞，，，，，结果，就像***一样炸成渣渣。。。。

#pragma GCC optimize("O3")#include <fstream>#include <cstring>int t, n, k, x, cnt, ans, tot;int to[200002], next[200002], first[100001];bool vis[100001];inline void read (int &x) {x = 0;char c = getchar();while (!isdigit(c)) c = getchar();while (isdigit(c)) x = x * 10 + c - 48, c = getchar();}inline void put (int x) {cnt = 0;char c[10];do c[++cnt] = x % 10 + 48, x /= 10; while(x);while (cnt) putchar(c[cnt--]);putchar('\n');}inline void add (int x, int y) {to[++tot] = y;next[tot] = first[x];first[x]  = tot;}void dfs (int x, int fa) {for (int i = first[x]; i; i = next[i]) if (to[i] ^ fa) {dfs (to[i], x);if (!vis[to[i]] and !vis[x]) {if (cnt < k) ++ans;cnt += 2;vis[to[i]] = vis[x] = true;}}}int main () {read(t);while (t--) {memset (first, 0, sizeof(first));memset (vis, 0, sizeof(vis));tot = cnt = ans = 0;read(n);read(k);for (int i = 2; i <= n; i++) {read(x);add (x, i);add (i, x);}dfs (1, 0);put (cnt < k ? ans + k - cnt : ans);}return 0;}