cf的几道小题

1、E – Fridge

教训：做题的时候，没有想清楚问题，把问题复杂化了

#include <bits/stdc++.h>
#define int long long
using namespace std;

const int N = 1010;
string st;
int cnt[N], minn = N, pos;
struct node
{
    int num;
    int sum;
    bool operator<(const node &p) const
    {
        if (sum < p.sum)
            return true;
        else if (sum == p.sum && num < p.num)
            return true;
        return false;
    }
} a[N];

signed main()
{
    cin >> st;
    for (int i = 0; i < st.size(); i++)
    {
        int x = st[i] - '0';
        cnt[x]++;
    }
    for (int i = 1; i <= 9; i++)
    {
        if (cnt[i] == 0)
        {
            cout << i;
            return 0;
        }
    }
    for (int i = 0; i < 10; i++)
    {
        a[i].num = i;
        a[i].sum = cnt[i];
    }
    if (!a[0].sum)
    {
        cout << 10;
        return 0;
    }
    sort(a + 1, a + 10);
    if (a[0].sum < a[1].sum)
    {
        cout << 1;
        for (int i = 1; i <= a[0].sum + 1; i++)
        {
            cout << 0;
        }
    }
    else
    {
        for (int i = 1; i <= a[1].sum + 1; i++)
        {
            cout << a[1].num;
        }
    }
    return 0;
}

2、J – Secret Santa

教训：数据范围比较大，直接暴力会超时，并且推公式比较麻烦，优先考虑打表

思路：当n >= 9时，会达到极限，后面的概率值都是一样的。

#include<bits/stdc++.h>
#define int long long
using namespace std;

const int N  = 1010;
int a[N], num = 1, n;
double cnt[N];

signed main(){
    cin >> n;
    for(int i = 1; i <= 10; i++) a[i] = i;
    int ans = 0;
    for(int i = 1; i <= 10; i++){
        ans = 0;
        num = num * i;
        do{
            for(int j = 1; j <= i; j++){
                if(a[j] == j){
                    ans ++;
                    break;
                }
            }
        }while(next_permutation(a + 1, a + 1 + i));
        cnt[i] =  1.0 * ans / num;
    }
    if(n >= 10) cout <<fixed << setprecision(8) << cnt[10];
    else cout <<fixed << setprecision(8) << cnt[n];
}

3、CF1656C Make Equal With Mod

思路：如果想让所有的序列变成0，那么只需从序列最大的数开始取模，这样最后的结果一定是0

#include<bits/stdc++.h>
using namespace std;

const int N = 1e5 + 10;
int n, a[N], t;

signed main(){
    cin >> t;
    while(t --){
        cin >> n;
        set<int> s;
        for(int i = 1; i <= n; i++){
            cin >> a[i];
            s.insert(a[i]);
        }
        bool flag1 = false, flag2 = false;
        for(set<int>::iterator it = s.begin(); it != s.end(); it ++){
            if(*it == 0) flag1 = true;
            if(*it == 1) flag2 = true;
        }
        if(flag1 && flag2){
            cout <<"NO" << endl;
        }
        else{
            sort(a + 1,a + 1 + n);
            bool flag = true;
            for(int i =1 ; i < n; i++){
                if(a[i + 1] - a[i] == 1 && a[1] == 1) {
                    flag = false; break;
                }
            }
            if(!flag) cout << "NO" << endl;
            else cout <<"YES" << endl;
        }
    }
    return 0;
}

4、D – Down the Pyramid

思路：求可变化的数的上界和下界，学会区分上界和下界。

#include<bits/stdc++.h>
using namespace std;

const int N = 1e6 + 10;
int n, a[N], now, minn = 0, maxn = 0x3f3f3f3f;

signed main(){
    cin >> n;
    for(int i = 1; i <= n; i++){
        cin >> a[i];
    }
    for(int i = 1; i <= n; i++){
        now = a[i] - now;
        if(i % 2 == 1){//偶数减x，所以x有最大值,要取上界最小值
            maxn = min(maxn, now);
        }
        else{//奇数 + x，所以x有最小值
            minn = max(minn, -now);
        }
    }
    if(minn > maxn) cout <<"0";
    else cout << maxn - minn + 1;
    return 0;
}

原文地址：http://www.cnblogs.com/N-lim/p/16907000.html