C. Candies

PS：大概是又傻了，讀題啊。

#include<bits/stdc++.h>
#define ll long long
#define P pair<int,int>
#define pb push_back
#define lson root << 1
#define INF (int)2e9 + 7
#define maxn (int)1e5 + 7
#define rson root << 1 | 1
#define LINF (unsigned long long int)1e18
#define mem(arry, in) memset(arry, in, sizeof(arry))
using
 
 namespace std;

ll n;

bool check(ll mid){
    ll res = 0;
    ll tp = n;
    while(tp){
        if(tp < mid) break;

        res += mid;
        tp = tp - mid;
        tp = tp - tp / 10;
    }
    res += tp;
    if(res * 2 >= n) return true;
    return false;
}

int main()
{
    cin >> n;
    ll l 
 
= 1, r = n;
    for(int i = 1; i <= 100; i++){
        ll mid = (l + r) >> 1;
        if(check(mid)) r = mid;
        else l = mid + 1;
    }
    cout << l << endl;
    return 0;
}

D. Bishwock

PS：又FST了，讀題的鍋。從頭依次尋找合法的位置放就行了。因為所有能放的位置在貪心的過程中都會被找到，故一定是最優解。

#include<bits/stdc++.h>
#define
 
 ll long long
#define P pair<int,int>
#define pb push_back
#define lson root << 1
#define INF (int)2e9 + 7
#define maxn (int)1e5 + 7
#define rson root << 1 | 1
#define LINF (unsigned long long int)1e18
#define mem(arry, in) memset(arry, in, sizeof(arry))
using namespace std;

char mp[2][200];

int main()
{
    for(int i = 0; i < 2; i++) scanf("%s", mp[i]);
    int n = strlen(mp[0]);
    int res = 0;
    for(int i = 0; i < n - 1; ){
        int t = 0;
        if(mp[0][i] == ‘X‘) t++;
        if(mp[1][i] == ‘X‘) t++;
        if(mp[0][i + 1] == ‘X‘) t++;
        if(mp[1][i + 1] == ‘X‘) t++;
        if(t > 1){
            i++;
            continue;
        }
        else{
            i += 2;
            res++;
        }

        int tt = 0;
        if(i < n - 2){
            if(mp[0][i] == ‘X‘) tt++;
            if(mp[1][i] == ‘X‘) tt++;
            if(t == 0 && tt == 0){
                i += 1;
                res++;
            }
        }
    }
    cout << res << endl;
    return 0;
}

E. Bus Number

PS：去重排列的公式得知道。然後暴力枚舉，可以用10個循環，也可以深搜。寫搜索很考驗狀態的設計，還有考慮什麽時候終止，在哪裏計算答案等等。

PS：非本人原創
#include<bits/stdc++.h>
#define ll long long
#define P pair<int,int>
#define pb push_back
#define pop pop_back
#define lson root << 1
#define INF (int)2e9 + 7
#define maxn (int)1e5 + 7
#define rson root << 1 | 1
#define LINF (unsigned long long int)1e18
#define mem(arry, in) memset(arry, in, sizeof(arry))
using namespace std;

int a[20];
ll n, res, fac[20];

vector<int> d;      //記錄每個數使用了多少次

void Inite()
{
    res = 0;
    fac[0] = 1;
    for(int i = 1; i < 20; i++) fac[i] = fac[i - 1] * i;
}

ll C(ll x, ll y){
    return fac[x] / fac[y] / fac[x - y];
}

void DFS(int x){
    if(x == 10){
        ll s = 0, tp = 0;
        for(auto o : d) s += o;
        tp = fac[s];
        for(auto o : d) tp /= fac[o];

        if(!a[0]){
            res += tp;
            return;
        }
        else{
            for(int i = 1; i <= a[0]; i++) res += tp * C(s + i - 1, i);
            return;
        }
    }
    if(!a[x]) DFS(x + 1);
    for(int i = 1; i <= a[x]; i++){
        d.pb(i);
        DFS(x + 1);
        d.pop();
    }
}

int main()
{
    Inite();
    cin >> n;
    while(n) {
       a[n % 10]++;
       n /= 10;
    }
    DFS(1);
    cout << res << endl;
    return 0;
}

Codeforces Round #491 (Div. 2) — 賽後補題