Codeforces Round #491 (Div. 2) — 赛后补题

Posted zgglj-com

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了Codeforces Round #491 (Div. 2) — 赛后补题相关的知识,希望对你有一定的参考价值。

C. Candies

PS:大概是又傻了,读题啊。

#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,读题的锅。从头依次寻找合法的位置放就行了。因为所有能放的位置在贪心的过程中都会被找到,故一定是最优解。

#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。然后暴力枚举,可以用10个循环,也可以深搜。写搜索很考验状态的设计,还有考虑什么时候终止,在哪里计算答案等等。

PS:非本人原创
#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; //记录每个数使用了多少次 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; }

 


以上是关于Codeforces Round #491 (Div. 2) — 赛后补题的主要内容,如果未能解决你的问题,请参考以下文章

Codeforces Round #491 (Div. 2) E - Bus Number + 反思

Codeforces Round #436 E. Fire(背包dp+输出路径)

[ACM]Codeforces Round #534 (Div. 2)

CodeforcesCodeforces Round #491 (Div. 2) (Contest 991)

Codeforces Round #726 (Div. 2) B. Bad Boy(贪心)

Codeforces 491B. New York Hotel 最远曼哈顿距离