A.直接暴力即可

​
#include<bits/stdc++.h>
using namespace std;
const int N =2e5+10;
#define int long long
typedef long long LL;
typedef pair<int, int> PII;
int n,m;void solve(){string s;cin>>s;for(int i=1;i<s.size();i++){int x=s[i-1]-'0';int y=s[i]-'0';if(x<=y){cout<<"No\n";return ;}}cout<<"Yes\n";
}signed main(){cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);int t=1;// cin>>t;while(t--) solve();
}​

B.也是暴力枚举每个分数即可

#include<bits/stdc++.h>
using namespace std;
const int N =2e5+10;
#define int long long
typedef long long LL;
typedef pair<int, int> PII;
int n,m;
int a[N];
void solve(){cin>>n>>m;for(int i=1;i<n;i++){cin>>a[i];}int res=INT_MAX;for(int j=0;j<=100;j++){vector<int> b;for(int i=1;i<n;i++) b.push_back(a[i]);b.push_back(j);sort(b.begin(),b.end());int now=0;for(int i=1;i<b.size()-1;i++) now+=b[i];if(now>=m){res=min(res,j);}}if(res==INT_MAX){cout<<-1;return ;}cout<<res;
}signed main(){cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);int t=1;// cin>>t;while(t--) solve();
}

C.我二分+数位dp了

直接二分当前数的范围内有多少个 “321”的数的个数

import random
import sys
import os
import math
from collections import Counter, defaultdict, deque
from functools import lru_cache, reduce,cache
from itertools import accumulate, combinations, permutations
from heapq import nsmallest, nlargest, heapify, heappop, heappush
from io import BytesIO, IOBase
from copy import deepcopy
import threading
import bisectBUFSIZE = 4096class FastIO(IOBase):newlines = 0def __init__(self, file):self._fd = file.fileno()self.buffer = BytesIO()self.writable = "x" in file.mode or "r" not in file.modeself.write = self.buffer.write if self.writable else Nonedef read(self):while True:b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))if not b:breakptr = self.buffer.tell()self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)self.newlines = 0return self.buffer.read()def readline(self):while self.newlines == 0:b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))self.newlines = b.count(b"\n") + (not b)ptr = self.buffer.tell()self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)self.newlines -= 1return self.buffer.readline()def flush(self):if self.writable:os.write(self._fd, self.buffer.getvalue())self.buffer.truncate(0), self.buffer.seek(0)class IOWrapper(IOBase):def __init__(self, file):self.buffer = FastIO(file)self.flush = self.buffer.flushself.writable = self.buffer.writableself.write = lambda s: self.buffer.write(s.encode("ascii"))self.read = lambda: self.buffer.read().decode("ascii")self.readline = lambda: self.buffer.readline().decode("ascii")sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")def I():return input()def II():return int(input())def MI():return map(int, input().split())def LI():return list(input().split())def LII():return list(map(int, input().split()))def GMI():return map(lambda x: int(x) - 1, input().split())def LGMI():return list(map(lambda x: int(x) - 1, input().split()))def solve():k = II()@cachedef dfs(i: int, limit: bool, num: bool, pre: int,s:str) -> int:if i == len(s):return int(num)res = 0if not num: res = dfs(i + 1, False, False, 10,s)up = int(s[i]) if limit else 9di = 0if not num: di = 1for d in range(di, up + 1):if pre > d:res += dfs(i + 1, limit and d == up, True, d,s)return resl, r = 1, 10 ** 18while l < r:mid = (l + r) >> 1if dfs(0, True, False, 10,str(mid)) >= k:r = midelse:l = mid + 1print(l)if __name__ == '__main__':for _ in range(1):solve()

D.

每个ai都要乘每个bi

排序后双指针维护当前ai+bi<=k的最有边界即可

#include<bits/stdc++.h>
using namespace std;
const int N =2e5+10;
#define int long long
typedef long long LL;
typedef pair<int, int> PII;
int n,m,k;
int a[N],b[N];
void solve(){cin>>n>>m>>k;for(int i=1;i<=n;i++){cin>>a[i];    }for(int j=1;j<=m;j++) cin>>b[j];sort(a+1,a+1+n);sort(b+1,b+1+m);int res=0;int now=0;for(int j=1;j<=m;j++) now+=b[j];int idx=m;for(int i=1;i<=n;i++){while(idx>=1&&a[i]+b[idx]>=k){now-=b[idx];idx--;}//    cout<<idx<<"\n";res+=idx*a[i]+now+(m-idx)*k;}cout<<res;
}signed main(){cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);int t=1;// cin>>t;while(t--) solve();
}

E.

直接get函数就是找以当前x为根节点的子树下面距离为k的有多少个数，最大值是n

那么最左边界是x<<k，最右边界是min(n,l+num-1)，

然后往父节点跳即可，边跳边计算父节点的另一个儿子节点

#include<bits/stdc++.h>
using namespace std;
const int N =2e5+10,mod= 998244353;
#define int long long
typedef long long LL;
typedef pair<int, int> PII;void solve()
{int n,x,k;int res=0;auto get=[&](int x,int k,int n){if(k>=64) return 0ll;if(k<0) return 0ll;__int128 num=1ll<<k,l=x;l=l<<k;__int128 r=l+num-1;if(l>n) return 0ll;LL res=min(r,(__int128)n)-l+1;return res;};cin>>n>>x>>k;if(k){res+=get(x,k,n);}while(x>1&&k>0){k--;res+=get((x^1),k-1,n);x/=2;}if(k==0&&x){res++;}cout<<res<<"\n";
}signed main(){cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);int t=1;cin>>t;while(t--) solve();
}

F：退背包问题

添加就是正常的01背包

然后删除就是

维护g数组

状态是，不选当前球，体积为i的方案数

那么g[i]=f[i]-g[i-x]

考虑容斥原理

不选当前球体积为i=选+不选当前球体积为i-选当前球体积为i的方案数

=f[i]-g[i-x]

g[i-x]可以理解为选了这个球后，剩下的 i−x�−�就是由不选该球的方案数组成

#include<bits/stdc++.h>
using namespace std;
const int N =2e5+10,mod= 998244353;
#define int long long
typedef long long LL;
typedef pair<int, int> PII;
int n,m;
int a[N];
int dep[N];void solve()
{int q,k;cin>>q>>k;vector<int> f(k+1,0);f[0]=1;while(q--){string op;int x;cin>>op>>x;if(op[0]=='+'){for(int i=k;i>=x;i--){f[i]+=f[i-x];f[i]%=mod;}}else{//不选该球和为j的方案书//选和不选vector<int> g(k+1,0);for(int i=0;i<=k;i++){if(i<x){g[i]=f[i];}else{g[i]=f[i]-g[i-x];if(g[i]<0) g[i]+=mod;}}f.swap(g);}cout<<f[k]<<"\n";}
}signed main(){cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);int t=1;// cin>>t;while(t--) solve();
}