SMU Summer 2023 Contest Round 11(2022-2023 ACM-ICPC Nordic Collegiate Programming Contest (NCPC 2022))

SMU Summer 2023 Contest Round 11(2022-2023 ACM-ICPC Nordic Collegiate Programming Contest (NCPC 2022))

A. Ace Arbiter

用\(A\)和\(B\)代表\(Alice\)和\(Bob\),则轮到\(Alice\)发球时会给出\(A - B\)的比分,轮到\(Bob\)时给出\(B-A\)的比分,直接比较还是挺难的,但是如果统一的都是\(A-B\)或者\(B-A\)的比分,那我们只要去判断每一轮的\(Alice\)和\(Bob\)的比分是不是大于等于上一轮的分数,如果少了,说明这次记录是错误的,另外如果有一方到达11分了,那么比赛应该结束了,不可能还继续比赛,还有就是\(11-11\)这种比分是不可能出现的.

那么问题来了,我们怎么去统一这个比分呢?

题目已给出发球顺序\(A,B,B,A,A,B,B\dots\),可以看出,除了第一轮\(Alice\)单独发球外,其余都是四轮一循环,且每一轮必定有一方会加分,所以我们可以将分数减掉\(Alice\)那一次,然后去 \(\bmod 4\),得到\(0\)或者\(1\)就说明这次是\(Bod\)发球,这个时候我们将比分对调,然后按照上面说的去判断即可.

#include<bits/stdc++.h>
 
using namespace std;
 
int32_t main() {
    int n;
    cin >> n;
    vector<int> a(n + 1), b(n + 1);
    char op;
    for (int i = 1; i <= n; i++)
        cin >> a[i] >> op >> b[i];
 
    bool end = false;
    for(int i = 1;i <= n;i ++){
 
        int sorce = max(a[i] + b[i] - 1,0);
        if(sorce % 4 == 0 || sorce % 4 == 1)
            swap(a[i],b[i]);
 
        if(end && a[i] != a[i - 1] || end && b[i] != b[i - 1]|| a[i] == 11 && b[i] == 11){
            cout << "error " << i << '\n';
            return 0;
        }
 
        if(a[i] < a[i - 1] || b[i] < b[i - 1]){
            cout << "error " << i << '\n';
            return 0;
        }
        if(a[i] == 11 || b[i] == 11)
            end = true;
    }
    cout << "ok\n";
    return 0;
}
 

C. Coffee Cup Combo

每个1可以将后面两个数变成1,但是由0变化来的1不能影响后面的数,按题意模拟即可

#include<bits/stdc++.h>
 
using namespace std;
 
int main(){
    int n;
    string s;
    cin >> n >> s;
    string sss = s;
    for(int i = 0;i < n;i ++){
        if(s[i] == '1'){
            sss[i] = '1';
            if(i + 1 < n) sss[i + 1] = '1';
            if(i + 2 < n) sss[i + 2] = '1';
        }
    }
 
    int ans = 0;
    for(auto i : sss)
        ans += (i == '1');
    cout << ans << '\n';
 
    return 0;
}
 

D. Disc District

多写几组数据可发现,其实\((r,1)\)也是符合答案

#include<bits/stdc++.h>
 
using namespace std;
 
int main() {
    long long r;
    cin >> r;
    cout << r << ' ' << 1 << endl;
    return 0;
}
 

G. Graduation Guarantee(期望dp)

先将概率从大到小排序(肯定要先做概率大的才能拿分嘛),然后就是\(dp\)数组的初始化,初始就是前面\(i\)道题做错.

你要想在前\(i\)道题里做对\(j\)道题,要么就是做对第\(j\)题,那么你前\((i-1)\)道题就要做对\((j-1)\)道题,要么就是不做第\(j\)道题,那你前\((i-1)\)道题就要做对\(j\)道题,由此得出转移方程

​ \(dp[i][j] = p[i] \times dp[i-1][j-1] + (1 - p[i]) \times dp[i-1][j]\)

\(i - (i-k)/2\)是说你在做对超出\(k\)道题时需要做错相应的题使你的总对题数维持在\(k\)

#include<bits/stdc++.h>
 
using namespace std;
 
int main() {
    
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
 
    int n,k;
    cin >> n >> k;
    std::vector<double> p(n + 1);
    for(int i = 1;i <= n;i ++)
        cin >> p[i];
 
    vector dp(n + 1, vector<double>(n + 1,0));
    sort(p.begin() + 1, p.end(),greater());
    dp[0][0] = 1;
    for (int i = 1; i <= n; ++i){
        dp[i][0] = (1 - p[i]) * dp[i - 1][0];
    }
 
    for(int i = 1;i <= n;i ++)
        for(int j = 1;j <= n;j ++)
            dp[i][j] = p[i] * dp[i - 1][j - 1] + (1 - p[i]) * dp[i - 1][j];
 
    double ans = 0.0;
    for(int i = k;i <= n;i ++){
        double sum = 0.0;
        for(int j = i - (i - k) / 2;j <= i;j ++)
            sum += dp[i][j];
 
        ans = max(ans, sum);
    }        
 
    cout << ans << '\n';
    return 0;
}
 

H. Highest Hill

思路就是每次一个峰一个峰地去找,然后更新答案.

思路还是挺对的,就是刚开始题目里有个等距导致我读了个假题,然后就是wa了几发,啊我真该死啊,最后贴份队友代码吧

#include<bits/stdc++.h>
#define int long long
#define endl '\n'
 
using namespace std;
 
int32_t main() {
    int n;
    cin >> n;
    vector<int> a(n + 10);
    a[n] = LLONG_MAX;
    for (int i = 0; i < n; i++)cin >> a[i];
    int a1, a2, a3;
    int pos1, pos2, pos3;
    int ans = 0;
    for (int i = 0; i < n; i++) {
        pos1 = i;
        while (a[i] <= a[i + 1] && i < n)i++;
        pos2 = i;
        if (pos1 == pos2)continue;
        while (a[i] >= a[i + 1] && i < n)i++;
        pos3 = i;
        if (i >= n)break;
        i--;
        
        a1 = a[pos1];
        a2 = a[pos2];
        a3 = a[pos3];
        ans = max(ans, min(a2 - a1, a2 - a3));
    }
    cout << ans << endl;
    return 0;
}