A - Long Comparison
B - Absent Remainder
C - Poisoned Dagger
D - MEX Sequences
E - Crazy Robot

 拿字符串比较

#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <cmath>
#include <stack>
#include <set>
#define mid (l+r>>1)
#define lowbit(x) (x&-x)
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
const int N = 2e6+10, mod = 1e9 + 7;
void mull(int &a, LL b){a = a*b%mod;return ;}
void add(int &a, LL b){a = (a+b)%mod;return ;}int main() 
{	int t;scanf("%d", &t);while(t --){string a, b;int c, d, f = 1;cin>>a>>c>>b>>d;while(a.size()<b.size()&&c)a+='0', c--;while(a.size()>b.size()&&d)b+='0', d--;int la = a.size()+c, lb = b.size()+d;if(la < lb) f = -1;else if(la == lb){if(a == b)f = 0;else if(a > b)f = 1;else f = -1;}if(f == 1)puts(">");else if(f == -1)puts("<");else puts("=");} return 0;
}

 $ymodx=t,ify>x,t<x$，那么x取最小值

#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <cmath>
#include <stack>
#include <set>
#define mid (l+r>>1)
#define lowbit(x) (x&-x)
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
const int N = 2e6+10, mod = 1e9 + 7;
void mull(int &a, LL b){a = a*b%mod;return ;}
void add(int &a, LL b){a = (a+b)%mod;return ;}int a[N];
int main() 
{	int t;scanf("%d", &t);while(t --){int n;scanf("%d", &n);for(int i = 0;i < n;i ++)scanf("%d", a+i);sort(a, a+n);int len = n/2, x = 1;while(len --)printf("%d %d\n", a[x++], a[0]);	} return 0;
}

 答案具有单调性，那么可以二分$nlogn$

#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <cmath>
#include <stack>
#include <set>
#define mid (l+r>>1)
#define lowbit(x) (x&-x)
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
const int N = 2e6+10, mod = 1e9 + 7;
void mull(int &a, LL b){a = a*b%mod;return ;}
void add(int &a, LL b){a = (a+b)%mod;return ;}int a[N], n;
LL h;bool ch(LL x)
{LL sum = 0, tail = 0;for(int i = 1;i <= n;tail = max(tail, a[i]+x), i ++)tail < a[i] ? sum += x : sum += a[i]+x-tail;return sum >= h;
}int main() 
{	int t;scanf("%d", &t);while(t --){scanf("%d%lld", &n, &h);for(int i = 1;i <= n;i ++)scanf("%d", a+i);LL l = 1, r = 1e18;while(l < r){if(ch(mid))r = mid;else l = mid+1;}cout<<l<<endl;} return 0;
}

 $dp,dp[i][0/1]表示mex=i，max=i-1ori+1$ 然后开始疯狂的推式子了，$a_i会和dp[a_i-1],dp[a_i],dp[a_i+1]$ 有影响，我把所有的数字都右移了1，因为不移1的话初始状态是$dp[-1][1]=1$

#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <cmath>
#include <stack>
#include <set>
#define mid (l+r>>1)
#define lowbit(x) (x&-x)
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
const int N = 2e6+10, mod = 998244353;
void mull(int &a, LL b){a = a*b%mod;return ;}
void add(int &a, LL b){a = (a+b)%mod;return ;}int in[N], dp[N][2];int main() 
{	int t;scanf("%d", &t);while(t --){int n;scanf("%d", &n);int ans = 0;dp[1][0] = 1;for(int i = 1;i <= n;i ++){int x;scanf("%d", &x); ++x;add(ans, (LL)dp[x+1][0]+dp[x+1][1]+dp[x][0]+dp[x-1][1]+dp[x-1][0]);add(dp[x-1][1], dp[x-1][1] + dp[x-1][0]);mull(dp[x+1][1], 2);add(dp[x+1][0], dp[x+1][0] + dp[x][0]);}for(int i = 1;i <= n+1;i ++)dp[i][0] = dp[i][1] = 0; cout<<ans<<endl;} return 0;
}

 这题想简单了，哎。要边遍历边修改

#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <cmath>
#include <stack>
#include <queue>
#define mid (l+r>>1)
#define lowbit(x) (x&-x)
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
const int N = 1e6+10, mod = 998244353;
void mull(int &a, LL b){a = a*b%mod;return ;}
void add(int &a, LL b){a = (a+b)%mod;return ;}int dx[4] = {1, -1, 0, 0}, dy[4] = {0, 0, 1, -1};int main() 
{	int t;scanf("%d", &t);while(t --){int n, m, x, y;scanf("%d%d", &n, &m);char a[n+1][m+1] ;   for(int i = 1;i <= n;i ++){scanf("%s", a[i]+1);for(int j = 1;j <= m;j ++)if(a[i][j] == 'L')x = i, y = j;}queue<PII>q;while(!q.empty())q.pop();	for(int i = 0;i < 4;i ++)q.push({x+dx[i], y+dy[i]});a[x][y] = '+';while(!q.empty()){int x = q.front().first, y = q.front().second, l = 0, r = 0;q.pop();if(!x || x>n || !y || y>m || a[x][y] != '.') continue;for(int i = 0;i < 4;i ++){int ll = x+dx[i], rr = y+dy[i];if(!ll || ll>n ||!rr || rr>m)continue;l += a[x+dx[i]][y+dy[i]] == '.', r += a[x+dx[i]][y+dy[i]] == '+';}if(!r || l > 1 )continue;a[x][y] = '+';for(int i = 0;i < 4;i ++)q.push({x+dx[i], y+dy[i]});}		a[x][y] = 'L';for(int i = 1;i <= n;i ++)printf("%s\n",a[i]+1);} return 0;
}
//#....
//..##L
//...#.
//.....