codeforces round 348 div2

最新推荐文章于 2025-07-28 11:16:14 发布

原创最新推荐文章于 2025-07-28 11:16:14 发布 · 362 阅读

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只会A、B、C

A 意思就是给出一个数，问你可以分成多少个3，因为3 = 1 + 2,这样交替给就行了，有余数再加1就解决。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>

using namespace std;
const int MAX = 505;

inline void file()
{
    freopen("D:\\zz.txt", "r", stdin);
    freopen("D:\\out.txt", "w", stdout);
}

int main()
{
    //file();
    int n;
    while (scanf("%d", &n) != EOF)
    {
        int res = 0;
        res = (n / 3) * 2 + ((n % 3 == 0) ? 0 : 1);
        printf("%d\n", res);
    }
    return 0;
}

B 当成图论判环来理解，重复走过某点是INFINITE。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <map>

using namespace std;
const int MAX = 505;
map<int, int> cnt;

inline void file()
{
    freopen("D:\\zz.txt", "r", stdin);
    freopen("D:\\out.txt", "w", stdout);
}

int main()
{
    //file();
    int n;
    char str[100005];
    int num[100005];
    while (scanf("%d", &n) != EOF)
    {
        memset(num, 0, sizeof(num));
        cnt.clear();
        scanf("%s", str);
        for (int i = 0; i < n; ++i)
            scanf("%d", &num[i]);
        int res = 0;
        bool vis = false;
        bool flag = false;
        for (int i = 0; i < n;)
        {
            if (str[i] == '<')
            {
                i -= num[i];
                cnt[i]++;
                if (i < 0 || i > n - 1)
                {
                    flag = true;
                    break;
                }
                if (cnt[i] > 1)
                {
                    vis = true;
                    break;
                }
            }
            else if (str[i] == '>')
            {
                i += num[i];
                cnt[i]++;
                if (i < 0 || i > n - 1)
                {
                    flag = true;
                    break;
                }
                if (cnt[i] > 1)
                {
                    vis = true;
                    break;
                }
            }
        }
        if (flag)
            printf("FINITE\n");
        else if (vis)
            printf("INFINITE\n");
    }
    return 0;
}

C 就是让你还原那个矩阵，先记住原来的位置，用一个数组存原来的位置，对这个数组进行变换，修改数字的时候，用原来的位置就行了。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <map>

using namespace std;
const int MAX = 505;
//vector<int> t[5];
int arr[105][105];
int n, m, q;
struct NODE
{
    int t_x, t_y;
    NODE(int x = 0, int y = 0)
    : t_x(x), t_y(y) {}
}node[105][105];

inline void file()
{
    freopen("D:\\zz.txt", "r", stdin);
    freopen("D:\\out.txt", "w", stdout);
}

void swap_x(int t)
{
    NODE tmp = node[t][1];
    for (int i = 1; i <= m - 1; ++i)
    {
        node[t][i] = node[t][i + 1];
    }
    node[t][m] = tmp;
}

void swap_y(int t)
{
    NODE tmp = node[1][t];
    for (int i = 1; i <= n - 1; ++i)
    {
        node[i][t] = node[i + 1][t];
    }
    node[n][t] = tmp;
}

int main()
{
    //file();
    scanf("%d%d%d", &n, &m, &q);
    for (int i = 1; i <= n; ++i)
    {
        for (int j = 1; j <= m; ++j)
        {
            node[i][j] = NODE(i, j);
        }
    }
    int type, pos;
    int tmp_x, tmp_y, tmp_val;
    while (q--)
    {
        scanf("%d", &type);
        if (type == 1)
        {
            scanf("%d", &pos);
            swap_x(pos);
            continue;
        }
        else if (type == 2)
        {
            scanf("%d", &pos);
            swap_y(pos);
            continue;
        }
        else if (type == 3)
        {
            scanf("%d%d%d", &tmp_x, &tmp_y, &tmp_val);
            arr[node[tmp_x][tmp_y].t_x][node[tmp_x][tmp_y].t_y] = tmp_val;
        }
    }
    for (int i = 1; i <= n; ++i)
    {
        for (int j = 1; j <= m; ++j)
        {
            printf("%d ", arr[i][j]);
        }
        printf("\n");
    }
    return 0;
}