個人題解連結，藍橋杯歷屆試題，正在更新中~

最優包含

題目連結：點選
思路：顯然的dp.
$設狀態為dp[i][j]代表前i個包含前j個要改變的次數$
轉移方程:
$ifs[i]==t[j]:dp[i][j]=dp[i-1][j-1]$
$else:dp[i][j]=min(dp[i-1][j],dp[i-1][j-1]+1)$
注意初始化為正無窮，$dp[i][0]$初始化為0

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int INF = 0x3f3f3f3f;
const double Pi = acos(-1);
namespace {
  template <typename T> inline void read(T &x) {
    x = 0; T f = 1;char s = getchar();
    for(; !isdigit(s); s = getchar()) if(s == '-') f = -1;
    for(;  isdigit(s); s = getchar()) x = (x << 3) + (x << 1) + (s ^ 48);
    x *= f;
  }
}
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define _for(n,m,i) for (register int i = (n); i <  (m); ++i)
#define _rep(n,m,i) for (register int i = (n); i <= (m); ++i)
#define _srep(n,m,i)for (register int i = (n); i >= (m); i--)
#define _sfor(n,m,i)for (register int i = (n); i >  (m); i--)
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define lowbit(x) x & (-x)
#define pii pair<int,int>
#define fi first
#define se second
const int N = 1e3+5;
char s[N], t[N];
int dp[N][N];// 前i個和包括了前j個要改幾次 
int main() {
  scanf("%s %s", s + 1, t + 1);
  int n = strlen(s+1), m = strlen(t+1);
  memset(dp, 0x3f, sizeof dp);
  
  dp[0][0] = 0;
  
  for(int i = 1; i <= n; ++i) {
    dp[i][0] = 0;
    for(int j = 1; j <= m && j <= i; ++j) {
      if(s[i] == t[j]) {
        dp[i][j] = dp[i-1][j-1];
      } else {
          dp[i][j] = min(dp[i-1][j], dp[i-1][j-1] + 1);
      }
    }
  }
  cout << dp[n][m] << endl;
}

排列數

題目連結：排列數
思路：顯然dp
$設狀態為dp[i][j]代表前i個數，有j個折點$
我們可以發現折點的數量就是波峰+波谷的數量。
雖然推出來後轉移方程很簡單，但是並不是特別的好推，我們得先把各種情況先看一遍，最後總結起來你就會發現轉移很簡單。
對於不同的折點數量，我們分為奇數偶數考慮，再對於每種情況我們分最後一個是波峰還是波谷考慮。
然後對於i個數，要變成i+1個數，肯定是插入了i+1，然後它有i+1個位置可以選擇。我們發現波峰周圍的2個位置是特殊的，在這裡插入，折點的數量不變。如果最後一個是波谷，那麼在最後插入會不變，如果第一個也是波谷，那麼在第一個位置插入也會不變（分成奇偶，和最後一個是波峰波谷就可以判斷出這兩種情況）。最後總結起來就是
$dp[i+1][j]+=dp[i][j]\ast 2$
$dp[i+1][j+1]=dp[i][j]\ast (j+1)$
$dp[i+1][j+2]=dp[i][j]\ast (i+1-j-1-2)$
答案就是$dp[n][k-1]$

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int INF = 0x3f3f3f3f;
const double Pi = acos(-1);
namespace {
  template <typename T> inline void read(T &x) {
    x = 0; T f = 1;char s = getchar();
    for(; !isdigit(s); s = getchar()) if(s == '-') f = -1;
    for(;  isdigit(s); s = getchar()) x = (x << 3) + (x << 1) + (s ^ 48);
    x *= f;
  }
}
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define _for(n,m,i) for (register int i = (n); i <  (m); ++i)
#define _rep(n,m,i) for (register int i = (n); i <= (m); ++i)
#define _srep(n,m,i)for (register int i = (n); i >= (m); i--)
#define _sfor(n,m,i)for (register int i = (n); i >  (m); i--)
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define lowbit(x) x & (-x)
#define pii pair<int,int>
#define fi first
#define se second
const int Mod = 123456;
const int N = 5e2+5;
int dp[N][N] ;
int main() {
  int n, k; read(n); read(k);
  dp[1][0] = 1;
  for(int i = 2;  i <= n; ++i) {
    dp[i][0] = 2;
    for(int j = 0; j <= i - 2; ++j) {
      dp[i][j] %= Mod;
      dp[i+1][j] += dp[i][j] * (j+1);
      dp[i+1][j+1] += dp[i][j] * 2;
      dp[i+1][j+2] += dp[i][j] * (i-j-2);
    }
  }
  cout << dp[n][k-1] % Mod << endl;
} 

第八大奇蹟

題目連結：點選
思路：這題其實就是一道裸題，整體二分or樹套樹，當然用整體二分好寫點。

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int INF = 0x3f3f3f3f;
const double Pi = acos(-1);
namespace {
  template <typename T> inline void read(T &x) {
    x = 0; T f = 1;char s = getchar();
    for(; !isdigit(s); s = getchar()) if(s == '-') f = -1;
    for(;  isdigit(s); s = getchar()) x = (x << 3) + (x << 1) + (s ^ 48);
    x *= f;
  }
}
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define _for(n,m,i) for (register int i = (n); i <  (m); ++i)
#define _rep(n,m,i) for (register int i = (n); i <= (m); ++i)
#define _srep(n,m,i)for (register int i = (n); i >= (m); i--)
#define _sfor(n,m,i)for (register int i = (n); i >  (m); i--)
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define lowbit(x) x & (-x)
#define pii pair<int,int>
#define fi first
#define se second
const int N = 1e5+5;
int a[N];
struct node {
  int op, x, y, k, id;
}e[N*10], lq[N], rq[N];
int cnt, ans[N];
int T[N];
int n, L;
void upd(int pos, int val) {
  for(; pos <= L; pos += lowbit(pos)) T[pos] += val;
}
int qry(int pos) {
  int res = 0;
  for(; pos; pos -= lowbit(pos)) res += T[pos];
  return res;
}
void solve(int vl, int vr, int ql, int qr) {
  if(ql > qr || vl > vr) return ;
  if(vl == vr) {
    for(int i = ql; i <= qr; ++i) {
      if(e[i].op == 2) ans[e[i].id] = vl;
    }
    return ;
  }
  int mid = vl + vr >> 1, l = 0, r = 0;
  for(int i = ql; i <= qr; ++i) {
    if(e[i].op == 1) {
      if(e[i].y > mid) {
        upd(e[i].x, e[i].k);
        rq[++r] = e[i];
      } else {
        lq[++l] = e[i];
      }
    } else {
      int k = qry(e[i].y) - qry(e[i].x-1);
      if(k < e[i].k) {
        e[i].k -= k;
        lq[++l] = e[i];        
      } else {
        rq[++r] = e[i];
      }
    }
  }
  for(int i = ql; i <= qr; ++i) {
    if(e[i].op == 1 && e[i].y > mid) {
      upd(e[i].x, -e[i].k);
    }
  }
  for(int i = 1; i <= l; ++i) e[i+ql-1] =  lq[i];
  for(int i = 1; i <= r; ++i) e[i+ql+l-1] = rq[i];
  solve(vl, mid, ql, ql+l-1);
  solve(mid+1, vr, ql+l, qr);
  
}
int main() {
  read(L); read(n);
  char op[3];
  memset(ans, -1, sizeof ans);
  for(int i = 1, x, y; i <= n; i++) {
    scanf("%s %d %d", op, &x, &y);
    if(op[0] == 'C') {
      if(a[x]) {
        e[++cnt] = node {1, x, a[x], -1};
        e[++cnt] = node {1, x, y, 1};
        a[x] = y;
      } else {
        e[++cnt] = node {1, x, y, 1};
        a[x] = y;
      }
    } else {
      e[++cnt] = node {2, x, y, 8, i};
    }
  }
  solve(0, 1e9, 1, cnt);
  for(int i = 1; i <= n; ++i) {
    if(~ans[i]) {
      printf("%d\n", ans[i]);
    }
  }
}