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

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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;}

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