P3879 [TJOI2010] 阅读理解 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

trie树板子题，稍微有一丢丢不一样，套用字典树模板稍加修改就能过

手搓字典树代码：

char ch[1010][26], cnt[1010], idx;
void insert(string s)//插入
{int p = 0;for (int i = 1; s[i]; i++){int j = s[i] - 'a';if (!ch[p][j]){ch[p][j] = ++idx;}p = ch[p][j];}cnt[p]++;
}
int query(string s)//查询
{int p = 0;for (int i = 1; s[i]; i++){int j = s[i] - 'a';if (!ch[p][j]){return 0;}p = ch[p][j];}return cnt[p];
}

题解代码：

#include<bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
#include<math.h>
using namespace std;
typedef long long ll;
char s[50010];
char ch[500010][26], idx;
char w[500010][1010];
int n, m;
inline int read()
{int k = 0, f = 1; char ch = getchar();while (ch < '0' || ch>'9'){if (ch == '-')f = -1;ch = getchar();}while (ch >= '0' && ch <= '9'){k = k * 10 + ch - '0';ch = getchar();}return k * f;
}
void insert(int x)
{scanf("%s", s + 1);int l = strlen(s + 1);int p = 0;for (int i = 1; i<=l; i++){int j = s[i] - 'a';if (!ch[p][j]){ch[p][j] = idx++;}p = ch[p][j];}w[p][x] = 1;
}
void check()
{scanf("%s", s + 1);int l = strlen(s + 1);int p = 0;int flag = 1;for (int i = 1; i<=l; i++){int j = s[i] - 'a';if (!ch[p][j]){flag = 0;break;}p = ch[p][j];}if (flag){for (int i = 1; i <= n; i++){if (w[p][i]){cout << i << " ";}}cout << endl;}
}
int main()
{ios::sync_with_stdio(false);cin.tie(nullptr);n = read();for (int i = 1; i <= n; i++){int x = read();for (int j = 1; j <= x; j++){insert(i);}}m = read();for (int i = 1; i <= m; i++){check();}return 0;
}

位运算：

左移：

1<<n=2^n,n<<1=2n

右移:

n>>1=n/2

取出n在二进制下的第k位：（n>>k）&1

取出n在二进制下的第0~k-1位（后k位）：n&((1<<k)-1)

把n在二进制下的第k位取反:n xor(1<<k)

对n在二进制下的第k位赋值1：n|(1<<k)

对n在二进制下的第k位赋值0:n&(~(1<<k))

来两道位运算的题目练练手（算法竞赛进阶指南）

P1226 【模板】快速幂 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

思路：

解决这道题有两种方法：

1.分治思想：n为偶数时把2^n分成a^(n/2),n为奇数时分为a^(n/2)*a

2.倍增思想:稍微有一点难理解(

从头开始。若当前 p 为偶数，咱们不着急，只需把 x 自乘，然后 p/=2 （即考虑下一层，下几层会帮我们乘上 (x2)p/2的）。

若当前 p 为奇数，说明 p=x∗(x2)(p−1)/2 中前面那个 x 的存在，ans∗=x。然后继续考虑下一层（下几层会帮我们乘上(x2)(p−1)/2的）

)

代码1（分治）：

#include<bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
#include<math.h>
#include <stdio.h>
using namespace std;
typedef long long ll;
inline int read()
{int k = 0, f = 1; char ch = getchar();while (ch < '0' || ch>'9'){if (ch == '-')f = -1;ch = getchar();}while (ch >= '0' && ch <= '9'){k = k * 10 + ch - '0';ch = getchar();}return k * f;
}
int solve(ll a, ll b, ll p)
{if (b == 0)return 1;ll ans = solve(a, b / 2, p) % p;if (b % 2 == 0){return ans * ans % p;}else{return ans * ans % p * a % p;}
}
int main()
{ios::sync_with_stdio(false);cin.tie(nullptr);ll a, b, p;cin >> a >> b >> p;cout << a << "^" << b << " mod " << p << "=" << solve(a, b, p) << endl;return 0;
}

代码2（倍增）：

#include<bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
#include<math.h>
#include <stdio.h>
using namespace std;
typedef long long ll;
inline int read()
{int k = 0, f = 1; char ch = getchar();while (ch < '0' || ch>'9'){if (ch == '-')f = -1;ch = getchar();}while (ch >= '0' && ch <= '9'){k = k * 10 + ch - '0';ch = getchar();}return k * f;
}
void solve(ll x, ll y, ll p)
{ll ans = 1 % p;for (; y; y >>= 1){if (y & 1){ans = ans * x % p;}x = x * x % p;}cout << x << "^" << y << " mod " << p << "=" << ans << endl;
}
int main()
{ios::sync_with_stdio(false);cin.tie(nullptr);ll a, b, p;cin >> a >> b >> p;solve(a, b, p);return 0;
}

P10446 64位整数乘法 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

代码：

#include<bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
#include<math.h>
#include <stdio.h>
using namespace std;
typedef long long ll;
inline int read()
{int k = 0, f = 1; char ch = getchar();while (ch < '0' || ch>'9'){if (ch == '-')f = -1;ch = getchar();}while (ch >= '0' && ch <= '9'){k = k * 10 + ch - '0';ch = getchar();}return k * f;
}
void solve(ll x, ll y, ll z)
{ll ans = 0;for (; y; y >>= 1){if (y & 1)ans = (ans + x) % 5;x = (x * 2) % z;}cout << ans;
}
int main()
{ios::sync_with_stdio(false);cin.tie(nullptr);ll a, b, p;cin >> a >> b >> p;solve(a, b, p);return 0;
}