Codeforces Round #292 (Div. 2)

A. Drazil and Date

无算法，判断(s - (a + b)) % 2是否为零，若零，表示在s步内还能走向其他的地方并且回来
否则，都是No

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <map>
#include <vector>
#include <set>
using namespace std;

const int MAXN = 1e6 + 10;
const int INF = 0x3f3f3f3f;

int main(void)
{
    //freopen ("A.in", "r", stdin);

    long long a, b, s;

    while (~scanf ("%I64d%I64d%I64d", &a, &b, &s))
    {
        if (a < 0)  a = -a;
        if (b < 0)  b = -b;
        if (a + b <= s)
        {
            int x = s - (a + b);
            if (x % 2 == 0) puts ("Yes");
            else    puts ("No");
        }
        else    puts ("No");
    }

    return 0;
}

B. Drazil and His Happy Friends

无算法，标记和更新happy的人就行了
少写一个&！，导致runtime error

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <map>
#include <vector>
#include <set>
using namespace std;

const int MAXN = 1e6 + 10;
const int INF = 0x3f3f3f3f;
int a[110], b[110];

int main(void)
{
    //freopen ("B.in", "r", stdin);

    int n, m;
    scanf ("%d%d", &n, &m);

    memset (a, 0, sizeof (a));
    memset (b, 0, sizeof (b));

    int x, y, tmp, cnt = 0;
    scanf ("%d", &x);
    for (int i=0; i<x; ++i)
    {
        scanf ("%d", &tmp);
        a[tmp] = 1;
        cnt++;
    }
    scanf ("%d", &y);
    for (int i=0; i<y; ++i)
    {
        scanf ("%d", &tmp);
        b[tmp] = 1;
        cnt++;
    }

    for (int i=0; i<=m*n*2; ++i)
    {
        if (a[i%n] || b[i%m])
        {
            if (!a[i%n])    cnt++;
            if (!b[i%m])    cnt++;
            a[i%n] = b[i%m] = 1;
        }
    }
    (cnt == n + m) ? puts ("Yes") : puts ("No");

    return 0;
}

C. Drazil and Factorial

无算法，数学问题貌似就是对单个数字分解质因数，替换，然后sort排序就行了

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <map>
#include <vector>
#include <set>
using namespace std;

const int MAXN = 1e6 + 10;
const int INF = 0x3f3f3f3f;

string res[10] = {"", "", "2", "3", "322", "5", "53", "7", "7222", "7332"};

int main(void)
{
    //freopen ("C.in", "r", stdin);

    int n;
    while (cin >> n)
    {
        string s, ans;
        cin >> s;
        for (int i=0; i<s.size(); ++i)
        {
            ans += res[s[i] - '0'];
        }
        sort (ans.begin (), ans.end ());
        reverse (ans.begin (), ans.end ());

        cout << ans << endl;
    }

    return 0;
}