Codeforces Round #246 (Div. 2) A,B,C,D

Codeforces Round #246 (Div. 2) A,B,C,D

A.水題，輸出圖形

B.水題

C.概率題

/*
m, n
最大數為k的總數為  k^n - (k-1)^n
所以最大數為k的期望為  (k^n - (k-1)^n) / (m^n)

*/

#include
using namespace std;

int main()
{
    int n, m;
    int i, j;
    scanf("%d%d", &m, &n);
    double ans = m;
    for(i=1; i<=m-1; ++i)
    {
        double x = i*1.0/m;
        ans = ans - pow(x, n);
    }
    printf("%.12f\n", ans);
    return 0;
}

D. 狀態壓縮DP

#include
using namespace std;

int pri[] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53};
int dp[105][1<<16];
int p[105][1<<16];
int mask[60];  //mask[i]已二進制形式表示是否含有pri[j]
int a[105];

int main()
{
    int n, i, j, k;
    for(i=1; i<60; ++i)
        for(j=0; j<16; ++j)
        if(i%pri[j]==0)
            mask[i] |= 1<dp[i-1][j] + abs(a[i]-k))
            {
                dp[i][j|mask[k]] = dp[i-1][j] + abs(a[i]-k);
                p[i][j|mask[k]] = k;
            }
        }
    }

    vector ans;
    for(int i=n, m=min_element(dp[i], dp[i]+(1<<16)) - dp[i]; i>0;
        m ^= mask[p[i][m]], --i)
        ans.push_back(p[i][m]);
    for(i=1; i<=n; ++i)
        printf("%d ", ans[n-i]);
    return 0;
}