牛客周赛 Round 76题解（ABC）

比赛链接：牛客周赛 Round 76

A-小红出题

思路：模拟即可。

#include <bits/stdc++.h>
#include <unordered_set>
#define endl '\n'
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, n, a) for (int i = n; i >= a; i--)
#define LL long long
#define IOS                  \
    ios::sync_with_stdio(0); \
    cin.tie(0);              \
    cout.tie(0);
using namespace std;

int n;

int main()
{
    IOS;
    cin >> n;
    int a = (n / 7) * 15;
    int b = 0;
    if(n%7<=5)
        b = (n % 7) * 3;
    else
        b = 5 * 3;
    cout << a + b;
    return 0;
}

B-串串香

思路：贪心的思想，显然越短的子串出现的次数可能越多，而最短的子串就是单个的字符，故只需要统计出现最多的字符是什么即可。

#include <bits/stdc++.h>
#include <unordered_set>
#define endl '\n'
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, n, a) for (int i = n; i >= a; i--)
#define LL long long
#define IOS                  \
    ios::sync_with_stdio(0); \
    cin.tie(0);              \
    cout.tie(0);
using namespace std;

int n;
string s;
int num[100005];

int main()
{
    cin >> n >> s;
    rep(i,0,s.size()-1)
    {
        num[s[i] - 'a']++;
    }
    int maxn = 0;
    char ans;
    rep(i,0,25)
    {
        if(num[i]>maxn)
        {
            maxn = num[i];
            ans = (char)('a' + i);
        }
            
    }
    cout << ans;
    return 0;
}

C-小红的gcd

思路：我们知道gcd(a,b)<=min(a,b)，贪心地思考，任意次操作后数组的最小值，就等于整个数组所有数的gcd。记得开long long。

#include <bits/stdc++.h>
#include <unordered_set>
#define endl '\n'
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, n, a) for (int i = n; i >= a; i--)
#define LL long long
#define IOS                  \
    ios::sync_with_stdio(0); \
    cin.tie(0);              \
    cout.tie(0);
using namespace std;

int n;
int a[100005];

LL gcd(LL a, LL b)
{
    while (b != 0)
    {
        LL temp = b;
        b = a % b;
        a = temp;
    }
    return a;
}

int main()
{
    cin >> n;
    rep(i, 1, n)
    {
        cin >> a[i];
        
    }
    LL agcd = a[1];
    rep(i,1,n)
    {
        agcd = gcd(agcd, a[i]);
    }
    cout << agcd * n;
    return 0;
}