Codeforces Round #392 (Div. 2)

最新推荐文章于 2025-08-01 11:12:48 发布

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原文链接：http://www.cnblogs.com/scaugsh/p/6332103.html

本文解析了四道编程题目，包括寻找最大数并补全、字符串合法化、学生提问次数统计及n进制转十进制等，通过具体代码实现介绍了各题目的解题思路。

传送门

A题：给你n个数，让你找出里面最大的数字，并把其他数字补到最大的数值，问需要补多少。直接两遍for就行了。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <string>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#define X first
#define Y second
#define clr(u,v); memset(u,v,sizeof(u));
#define in() freopen("data","r",stdin);
#define out() freopen("ans","w",stdout);
#define Clear(Q); while (!Q.empty()) Q.pop();
#define pb push_back
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int maxn = 1e5 + 10;
const int INF = 0x3f3f3f3f;
int N[maxn];
int main()
{
    int n;
    scanf("%d", &n);
    int Max = -INF;
    for (int i = 0; i < n; i++)
    {
        cin >> N[i];
        Max = max(Max, N[i]);
    }
    ll ans = 0;
    for (int i = 0; i < n; i++)
        ans += Max - N[i];
    cout << ans << endl;
    return 0;

}

View Code

B题：给你一个字符串，包含RBYG!五种字符，其中!字符可以修改为任意字符，同时，字符串按照abcdabcd……这样的格式，abcd分别为RBYG中的任意一个，且互不相同。现在问你要分别把!改成哪些字符才能使原字符串合法，输出需要多少个RBYG。我们只要考虑前四位就行了，dfs出前四位的所有情况，然后判断是不是合法。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <string>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#define X first
#define Y second
#define clr(u,v); memset(u,v,sizeof(u));
#define in() freopen("data","r",stdin);
#define out() freopen("ans","w",stdout);
#define Clear(Q); while (!Q.empty()) Q.pop();
#define pb push_back
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int maxn = 1e5 + 10;
const int INF = 0x3f3f3f3f;
char str[210];
int R = 0, B = 0, Y = 0, G = 0;
int N[100];
const char s[] = {"RBYG"};
bool ok(int cur, int n)
{
    for (int i = 0; i < cur; i++)
        if (str[i] == str[cur]) return 0;
    for (int i = cur + 4; i < n; i += 4) 
        if (str[i] != '!' && str[cur] != str[i]) return 0;
    for (int i = cur + 4; i < n; i += 4)
        if (str[i] == '!') N[str[cur]]++;
    return 1;
}

void dfs(int cur , int n, int len)
{
    if (cur == n)
        return ;
    if (str[cur] == '!')
    {
        for (int i = 0; i < 4; i++)
        {
            str[cur] = s[i];
            if (ok(cur, len))
            {
                N[s[i]]++;
                dfs(cur + 1, n, len);
                return ;
            }
        }
    }
    else ok(cur, len);
    dfs(cur + 1, n, len);
}

int main()
{
    scanf("%s", str);
    int len = strlen(str);
    dfs(0, 4, len);
    for (int i = 0; i < 4; i++)
        cout << N[s[i]] << " ";
    return 0;
}

View Code

C题：有n*m个学生，老师有k个问题，按照行:1~n~1,列1~m的顺序提问。然后给你Sergei的坐标，让你输出班上被问问题最多的次数，最少的次数以及Sergei被问了多少次。由于n和m不大，可以借助数组来模拟一下（不用数组的话细节很多）。可以考虑，(2n-1)*m是一个周期，所以直接模拟k%((2n-1)*m)这一部分就行了，复杂度不会很高，当然要注意特判n==1的情况。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <string>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#define X first
#define Y second
#define clr(u,v); memset(u,v,sizeof(u));
#define in() freopen("data","r",stdin);
#define out() freopen("ans","w",stdout);
#define Clear(Q); while (!Q.empty()) Q.pop();
#define pb push_back
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int maxn = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const ll inf = 1e18 + 10;
ll M[210][210];
int main()
{
    ll n, m, k;
    ll x, y;
    cin >> n >> m >> k >> x >> y;
    if (n == 1)
    {
        ll add = k / m, mod = k % m;
        for (int j = 1; j <= m; j++)
            M[1][j] += add;
        for (int j = 1; j <= m && mod; j++)
        {
            M[1][j]++;
            mod--;
        }
    }
    else
    {
        ll add = k / ((2 * n - 2) * m);
        for (int i = 2; i < n; i++)
            for (int j = 1; j <= m; j++)
                M[i][j] += 2 * add;
        for (int j = 1; j <= m; j++)
            M[n][j] += add;
        for (int j = 1; j <= m; j++)
            M[1][j] += add;
        ll mod = k % ((2 * n - 2) * m);
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= m; j++)
                if (mod)
                {
                    M[i][j]++;
                    mod--;
                }
                else break;
        for (int i = n - 1; i > 1; i--)
            for (int j = 1; j <= m; j++)
                if (mod)
                {
                    M[i][j]++;
                    mod--;
                }
                else break;
    }
    ll ans = M[x][y] ;
    ll Min = inf, Max = 0;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
        {
            if (M[i][j] != -1)
            {
                Min = min(Min, M[i][j]);
                Max = max(Max, M[i][j]);
            }
        }
    cout << Max << " " << Min << " " << ans;
    return 0;
}

View Code

D题：给你一个n进制的数字k，让你把它转为十进制，且这个数字ans要最小。因为会出现数字k中的两位或者多位组合成ans中的一位，所以ans不是唯一的。可以考虑贪心，从后面取数字，经可能取大的数字。大概证明：假设数字为ab，为c进制，那么有a<c,b<c,假设ab<c,那么ab*c<c*c<a*c*c+b*c，这就说明如果能选多位，尽可能选多位。这题细节很多，最麻烦的就是0的问题了，细节自己搞定吧。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <string>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#define X first
#define Y second
#define clr(u,v); memset(u,v,sizeof(u));
#define in() freopen("data","r",stdin);
#define out() freopen("ans","w",stdout);
#define Clear(Q); while (!Q.empty()) Q.pop();
#define pb push_back
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int maxn = 1e5 + 10;
const int INF = 0x3f3f3f3f;
char str[maxn];
ll pow(ll n, ll k)
{
    ll ans = 1;
    while (k > 0)
    {
        if (k & 1) ans *= n;
        k >>= 1;
        n = n * n;
    }
    return ans;
}
int main()
{
    ll n, ans = 0;
    cin >> n;
    cin >> (str + 1);
    int len = strlen(str + 1);
    ll k = 0;
    str[0] = '0';
    int pre = len;
    for (int i = len; i >= 0; i--)
    {
        int pos = i;
        ll num = 0;
        while (pos <= pre && num < n)
        {
            num *= 10;
            num += str[pos] - '0';
            pos++;
        }
        if (num >= n || i == 0)
        {
            pos = i + 1;
            num = 0;
            int flag = 0;
            while (str[pos] == '0')
            {
                pos++;
                flag = 1;
            }
            int temp = pos - 1;
            while (pos <= pre)
            {
                num *= 10;
                num += str[pos] - '0';
                pos++;
            }
            if (num == 0 && str[pre] == '0')
            {
                str[pre] = '\0';
                k++;
                i++;
                pre--;
                continue;
            }
            if (flag) i++;
            ans += num * pow(n, k);
            k++;
            pre = temp;
        }
    }
    cout << ans << endl;
    return 0;
}