A.水题，输出图形

B.水题

C.概率题

/*
m, n
最大数为k的总数为  k^n - (k-1)^n
所以最大数为k的期望为  (k^n - (k-1)^n) / (m^n)

*/

#include<bits/stdc++.h>
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. 状态压缩DP

#include<bits/stdc++.h>
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]已二进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<<j;
    scanf("%d", &n);
    for(i=1; i<=n; ++i) scanf("%d", &a[i]);
    memset(dp, 0x3f, sizeof dp );
    dp[0][0] = 0;
    for(i=1; i<=n; ++i)
    {
        for(j=0; j< (1<<16); ++j)
        for(k=1; k<60; ++k) if( (mask[k]&j)==0)
        {
            if(dp[i][j|mask[k]]>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<int> 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;
}