【NOIP模拟】20/02/15

• 
• 直接，，前缀和优化，只有个有用位置

#include<bits/stdc++.h>
#define cs const
using namespace std;
cs int Mod = 1e9 + 7;
int add(int a, int b){ return a + b >= Mod ? a + b - Mod : a + b; }
int mul(int a, int b){ return 1ll * a * b % Mod; }
int dec(int a, int b){ return a - b < 0 ? a - b + Mod : a - b; }
int ksm(int a, int b){ int as=1; for(;b;b>>=1,a=mul(a,a)) if(b&1) as=mul(as,a); return as; }
void Add(int &a, int b){ a = add(a, b); }
void Mul(int &a, int b){ a = mul(a, b); }
cs int N = 1e5 + 50;
int n, m, S, sum[N];
int ps1[N], ps2[N], nd, vl[N]; 
int idx(int x){ if(x <= S) return ps1[x]; return ps2[m/x]; }
int main(){
  freopen("noname.in","r",stdin);
  freopen("noname.out","w",stdout);
  cin >> n >> m; S = sqrt(m); int c = 0;
  for(int l = 1, r; l <= m; l = r+1) r = m / (m / l), ++c;
  nd = c;
  for(int l = 1, r; l <= m; l = r+1){
    int v = m / l; r = m / v;
    (v <= S) ? ps1[v] = c-- : ps2[m/v] = c--;
    sum[c+1] = vl[c+1] = v;
  } 
  for(int t = 1; t < n; t++){
    static int nxt[N]; 
    for(int i = 1; i <= nd; i++){
      nxt[i] = add(nxt[i-1], mul(vl[i]-vl[i-1], sum[idx(m/vl[i])]));
    } memcpy(sum, nxt, sizeof(int)*(nd+1));
  } cout << sum[nd]; return 0;
}

• 思路挺巧妙的，我开始想的是考虑一个数插头插尾的贡献用个线段树之类的维护（放后面是区间加，放前面是查询最大值），应该也可以
  正解比较妙，考虑将序列反过来赋值一遍拼在前面，那么就是前面选一个子序列后面选一个子序列子序列不相交，最大化，发现严格递增的一定不会相交，所以对新序列求一个就可以了

#include<bits/stdc++.h>
#define cs const
using namespace std;
int read(){
  int cnt = 0, f = 1; char ch = 0;
  while(!isdigit(ch)){ ch = getchar(); if(ch == '-') f = -1; }
  while(isdigit(ch)) cnt = cnt * 10 + (ch-'0'), ch = getchar();
  return cnt * f;
}
cs int N = 1e5 + 50;
int n, A[N]; set<int> S;
typedef set<int>::iterator It;
int main(){
  freopen("dequexlis.in","r",stdin);
  freopen("dequexlis.out","w",stdout);
  n = read();
  for(int i = 1; i <= n; i++) A[i] = read();
  for(int i = n; i >= 1; i--){
    It it = S.lower_bound(A[i]);
    if(it == S.end()) S.insert(A[i]);
    else S.erase(it), S.insert(A[i]);
  }
  for(int i = 1; i <= n; i++){
    It it = S.lower_bound(A[i]);
    if(it == S.end()) S.insert(A[i]);
    else S.erase(it), S.insert(A[i]);
  } cout << S.size(); return 0;
}

• 不是很难，但蛮有趣的，考虑 1 为向上走，-1 为向右走，那么贡献就是与相交的的最大值，考虑统计每一个有多少个，容斥一下求的那么方案数就是，跟卡特兰数差不多

#include<bits/stdc++.h>
#define cs const
using namespace std;
cs int N = 2e6 + 50;
cs int Mod = 998244853;
int add(int a, int b){ return a + b >= Mod ? a + b - Mod : a + b; }
int mul(int a, int b){ return 1ll * a * b % Mod; }
int dec(int a, int b){ return a - b < 0 ? a - b + Mod : a - b; }
int ksm(int a, int b){ int as=1; for(;b;b>>=1,a=mul(a,a)) if(b&1) as=mul(as,a); return as; }
void Add(int &a, int b){ a = add(a, b); }
void Mul(int &a, int b){ a = mul(a, b); }
int fix(int a){ return a < 0 ? -a : a; }
int n, m, fac[N], ifac[N];
int C(int n, int m){ if(n<0||m<0||n<m) return 0; return mul(fac[n],mul(ifac[n-m],ifac[m])); }
void prework(int n){
  fac[0] = fac[1] = ifac[0] = ifac[1] = 1;
  for(int i = 2; i <= n; i++) fac[i] = mul(fac[i-1], i);
  ifac[n] = ksm(fac[n], Mod-2);
  for(int i = n-1; i >= 2; i--) ifac[i] = mul(ifac[i+1], i+1);
}
int main(){
  freopen("maxpsum.in","r",stdin);
  freopen("maxpsum.out","w",stdout)
  cin >> n >> m; prework(n+m); int as = 0;
  for(int i = max(1,n-m); i <= n; i++) 
    Add(as, mul(i,dec(C(n+m, m+i),C(n+m, m+i+1))));
  cout << as; return 0; 
}