A. Div. 7

思路：
$注意n>=10，说明一定是两位数$
$并且7个一循环，直接枚举这位数的最后一位$
$暴力判断是否可行即可$
时间复杂度：$Ot$

#include <bits/stdc++.h>
#define fer(i,a,b) for(re i = a ; i <= b ; ++ i)
#define der(i,a,b) for(re i = a ; i >= b ; -- i)
#define cf int _; cin>> _; while(_--)
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define sf(x) scanf("%lld",&x)
#define pll pair<int,int> 
#define re register int
#define lb lower_bound
#define ub upper_bound
#define int long long 
#define pb push_back
#define y second 
#define x first 
using namespace std;
inline void sf2(int &a , int &b) { sf(a) , sf(b) ;}
inline void sf3(int n , int a[]) {fer(i,1,n) sf(a[i]);} ;
inline void de(auto x) {cout << x << "\n" ;}
inline void de2(auto a , auto b) {cout << a << " " << b << "\n" ;}
inline void de3(int n , auto a[]){fer(i,1,n) cout << a[i] << " " ;puts("");}
inline void mo(int &a , int b) {a = (a % b + b) % b ;}
inline int gcd(int a,int b){return b ? gcd(b , a % b) : a ;}
inline int qpow(int a,int b,int c){int res=1%c;a%=c;while(b>0){if(b&1)res=res*a%c;a=a*a%c;b>>=1;}return res;}
inline int qadd(int a,int b,int c){int res=0;a%=c;while(b>0){if(b&1)res=(res+a)%c;a=(a+a)%c;b>>=1;}return res;} 
const int inf = 0x3f3f3f3f3f3f3f3f3f3f3f3f3f3f3f3f ;
const int N = 1e6 + 10 , M = 3010 , mod = 1e9 + 7 ;
const double eps = 1e-7 , pi = acos(-1.0) ;

signed main()
{
    cf
    {
        int n ;
        cin >> n ;
        if(n % 7 == 0) de(n) ;
        else
        {
            string s = to_string(n) ;

            for(char i = '0' ; i <= '9' ; i ++)
            {
                s[sz(s) - 1] = i ;
                if(stoi(s) % 7 == 0)
                {
                    de(s) ;
                    break ;
                }
            }
        }
    }
    return 0;
}

B. Minority

思路：
$假设该字符串0的数量为a$
$1的数量为b$
$如果a=b，显然答案为max(a-1,0)$
$如果a>b，答案为b$
$如果a<b，答案为a$
时间复杂度：$On$

#include <bits/stdc++.h>
#define fer(i,a,b) for(re i = a ; i <= b ; ++ i)
#define der(i,a,b) for(re i = a ; i >= b ; -- i)
#define cf int _; cin>> _; while(_--)
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define sf(x) scanf("%lld",&x)
#define pll pair<int,int> 
#define re register int
#define lb lower_bound
#define ub upper_bound
#define int long long 
#define pb push_back
#define y second 
#define x first 
using namespace std;
inline void sf2(int &a , int &b) { sf(a) , sf(b) ;}
inline void sf3(int n , int a[]) {fer(i,1,n) sf(a[i]);} ;
inline void de(auto x) {cout << x << "\n" ;}
inline void de2(auto a , auto b) {cout << a << " " << b << "\n" ;}
inline void de3(int n , auto a[]){fer(i,1,n) cout << a[i] << " " ;puts("");}
inline void mo(int &a , int b) {a = (a % b + b) % b ;}
inline int gcd(int a,int b){return b ? gcd(b , a % b) : a ;}
inline int qpow(int a,int b,int c){int res=1%c;a%=c;while(b>0){if(b&1)res=res*a%c;a=a*a%c;b>>=1;}return res;}
inline int qadd(int a,int b,int c){int res=0;a%=c;while(b>0){if(b&1)res=(res+a)%c;a=(a+a)%c;b>>=1;}return res;} 
const int inf = 0x3f3f3f3f3f3f3f3f3f3f3f3f3f3f3f3f ;
const int N = 1e6 + 10 , M = 3010 , mod = 1e9 + 7 ;
const double eps = 1e-7 , pi = acos(-1.0) ;

signed main()
{
    cf
    {
        string s ;

        int s1 = 0 , s2 = 0 ;

        cin >> s ;

        for(auto i : s)
        {
            if(i == '1') s1 ++ ;
            else s2 ++ ;
        }

        if(s1 == s2) de(max(s1 - 1 , 0ll)) ;
        else if(s1 > s2) de(s2) ;
        else de(s1) ;
    }
    return 0;
}

C. Kill the Monster

思路：
$注意到k的总和为2e5$
$那么我们可以枚举生命hc加i次,攻击dc加k-i次$
$o1判断即可$
$时间复杂度ok$

$如何o1判断$
$敌人杀死我需要\lceil \frac{hc}{dm} \rceil个回合$
$我杀死敌人需要\lceil \frac{hm}{dc} \rceil个回合$
$即$
$$\lceil \frac{hc}{dm} \rceil >= \lceil \frac{hm}{dc} \rceil$$

时间复杂度：$Ok$

#include <bits/stdc++.h>
#define fer(i,a,b) for(re i = a ; i <= b ; ++ i)
#define der(i,a,b) for(re i = a ; i >= b ; -- i)
#define cf int _; cin>> _; while(_--)
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define sf(x) scanf("%lld",&x)
#define pll pair<int,int> 
#define re register int
#define lb lower_bound
#define ub upper_bound
#define int long long 
#define pb push_back
#define y second 
#define x first 
using namespace std;
inline void sf2(int &a , int &b) { sf(a) , sf(b) ;}
inline void sf3(int n , int a[]) {fer(i,1,n) sf(a[i]);} ;
inline void de(auto x) {cout << x << "\n" ;}
inline void de2(auto a , auto b) {cout << a << " " << b << "\n" ;}
inline void de3(int n , auto a[]){fer(i,1,n) cout << a[i] << " " ;puts("");}
inline void mo(int &a , int b) {a = (a % b + b) % b ;}
inline int gcd(int a,int b){return b ? gcd(b , a % b) : a ;}
inline int qpow(int a,int b,int c){int res=1%c;a%=c;while(b>0){if(b&1)res=res*a%c;a=a*a%c;b>>=1;}return res;}
inline int qadd(int a,int b,int c){int res=0;a%=c;while(b>0){if(b&1)res=(res+a)%c;a=(a+a)%c;b>>=1;}return res;} 
const int inf = 0x3f3f3f3f3f3f3f3f3f3f3f3f3f3f3f3f ;
const int N = 1e6 + 10 , M = 3010 , mod = 1e9 + 7 ;
const double eps = 1e-7 , pi = acos(-1.0) ;

bool check(int hc , int dc , int hm , int dm)
{
    int len = (hc + dm - 1) / dm ;
    if(len * dc >= hm) return 1 ;
    else return 0 ;
}

signed main()
{
    cf
    {
        int hc , dc , hm , dm , k , w , a ;

        sf2(hc , dc) ;
        sf2(hm , dm) ;
        sf(k) ;
        sf2(w , a) ;

        int f1 = 0 ;

        for(int i = 0 ; i <= k ; i ++)
        {
            int j = k - i ;

            if(check(hc + i * a , dc + j * w , hm , dm )) f1 = 1 ;    
        }

        if(f1) puts("YES") ;
        else puts("NO") ;
    }
    return 0;
}

D. Make Them Equal

思路：
$注意到每次a[i]变成b[i]都是从1开始变化的$

$也就是说我们可以先预处理一个数组d[i]数组$
$表示1变化到i的最小操作数$

$注意到b[i]的最大值是1e3$
$我们可以暴力bfs用O1e6的时间复杂度预处理该数组$

$然后在把题目转换一下$
$操作数最多是k$
$对于每个i,变成b[i]的操作数为d[i]$
$你可以得到c[i]个硬币$

$显然,操作数k我们可以看作是背包体积$
$对于每个i,变成b[i]的操作数为d[i],可以看作是物品i的体积为d[i]$
$你可以得到c[i]个硬币,可以看作是物品i的价值c[i]$

$每个物品最多选一次,跑01背包即可$

$注意到d[i]数组的最大值为12$
$也就是说背包体积最大为12*1e3$
$假设背包体积为m,m=min(12*n,k)$

$时间复杂度=min(12*n,k)*n + n^2 \approx 1e7$
时间复杂度：$Omin(12*n,k)*n + n^2$

#include <bits/stdc++.h>
#define fer(i,a,b) for(re i = a ; i <= b ; ++ i)
#define der(i,a,b) for(re i = a ; i >= b ; -- i)
#define cf int _; cin>> _; while(_--)
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define sf(x) scanf("%lld",&x)
#define pll pair<int,int> 
#define re register int
#define lb lower_bound
#define ub upper_bound
#define int long long 
#define pb push_back
#define y second 
#define x first 
using namespace std;
inline void sf2(int &a , int &b) { sf(a) , sf(b) ;}
inline void sf3(int n , int a[]) {fer(i,1,n) sf(a[i]);} ;
inline void de(auto x) {cout << x << "\n" ;}
inline void de2(auto a , auto b) {cout << a << " " << b << "\n" ;}
inline void de3(int n , auto a[]){fer(i,1,n) cout << a[i] << " " ;puts("");}
inline void mo(int &a , int b) {a = (a % b + b) % b ;}
inline int gcd(int a,int b){return b ? gcd(b , a % b) : a ;}
inline int qpow(int a,int b,int c){int res=1%c;a%=c;while(b>0){if(b&1)res=res*a%c;a=a*a%c;b>>=1;}return res;}
inline int qadd(int a,int b,int c){int res=0;a%=c;while(b>0){if(b&1)res=(res+a)%c;a=(a+a)%c;b>>=1;}return res;} 
const int inf = 0x3f3f3f3f3f3f3f3f3f3f3f3f3f3f3f3f ;
const int N = 1e6 + 10 , M = 3010 , mod = 1e9 + 7 ;
const double eps = 1e-7 , pi = acos(-1.0) ;

int n , k ;
int b[N] ;
int c[N] ;
int d[N] ;
int f[N] ;

signed main()
{
    // bfs预处理d数组
    memset(d , -1 , sizeof d) ;
    d[1] = 0 ;
    queue<int> q ;
    q.push(1) ;
    while(q.size())
    {
        auto t = q.front() ;
        q.pop() ;

        for(int j = 1 ; j <= t ; j ++)
        {
            int x = t + t / j ;
            if(x > 1e3) continue ;
            if(d[x] != -1 || x == 1) continue ;

            d[x] = d[t] + 1 ;
            q.push(x) ;
        }
    }

    cf
    {
        cin >> n >> k ;
        fer(i,1,n) sf(b[i]) ;
        fer(i,1,n) sf(c[i]) ;

        int m = min(k,20000*1ll) ;
        fer(i,0,m) f[i] = 0 ;

        // 01背包
        for(int i = 1 ; i <= n ; i ++)
        {
            for(int j = m ; j >= d[b[i]] ; j --)
                f[j] = max(f[j],f[j - d[b[i]]] + c[i]);
        }

        de(f[m]) ;
    }
    return 0;
}