第九届蓝桥杯C++B组题解

#include 
#include 
#include 

using namespace std;
typedef long long LL;

const int N = 110, mod = 1e9 + 7;

int n, m;
LL s[N][N], c[N][N];
LL f[N][N][N];

void get_prefix(int i)
{
    for (int j = 1; j <= m; j ++ )
        for (int k = 1; k <= m; k ++ )
            s[j][k] = (s[j - 1][k] + s[j][k - 1] - s[j - 1][k - 1] + f[i][j][k]) % mod;
}

LL get_sum(int x1, int y1, int x2, int y2)
{
    return s[x2][y2] - s[x2][y1 - 1] - s[x1 - 1][y2] + s[x1 - 1][y1 - 1];
}

int main()
{
    scanf("%d%d", &n, &m);
    
    char str[N];
    for (int i = n; i; i -- )
    {
        cin >> str + 1;
        for (int j = 1; j <= m; j ++ )
        {
            c[i][j] = c[i][j - 1] + (str[j] == 'X');
        }
    }
    
    f[0][1][m] = 1;
    get_prefix(0);
    
    LL res = 1;
    for (int i = 1; i <= n; i ++ )
    {
        for (int j = 1; j <= m; j ++ )
            for (int k = j; k <= m; k ++ )
            {
                if (c[i][k] - c[i][j - 1] == 0)
                {
                    LL &x = f[i][j][k];
                    x = (x + get_sum(1, k, j, m)) % mod;
                    res = (res + x) % mod;
                }
            }
        get_prefix(i);
    }
    
    printf("%lldn", (res + mod) % mod);
    return 0;
}

#include 
#include 
#include 

using namespace std;
typedef long long LL;
const int N = 1e7 + 9, mod = 1e9 + 7;
LL euler[N], primes[N];
bool st[N];
LL s[N];
int n, cnt;

void get_eulers(int n)  // 线性筛法求1~n的欧拉函数
{
    euler[1] = 1;
    for (int i = 2; i <= n; i ++ )
    {
        if (!st[i])
        {
            primes[cnt ++ ] = i;
            euler[i] = i - 1;
        }
        for (int j = 0; primes[j] <= n / i; j ++ )
        {
            int t = primes[j] * i;
            st[t] = true;
            if (i % primes[j] == 0)
            {
                euler[t] = euler[i] * primes[j];
                break;
            }
            euler[t] = euler[i] * (primes[j] - 1);
        }
    }
    
    s[1] = 1;
    for (int i = 2; i <= n; i ++ )
        s[i] = (s[i - 1] + 2 * euler[i]) % mod;
}

int main()
{
    cin >> n;
    
    get_eulers(n);
    
    LL res = 0;
    for (int d = 1; d <= n; d ++ )
        res = (res + s[n / d] * d % mod * d) % mod;
    printf("%lldn", res);
    return 0;
}