Hdu5159--bc--card (probability notation, and combinatorial notation)

Probability method

Demands out and expectations, the basic theorem of expectation, and the expectation of each part of the desired and.

E (sum) = e (1) + E (2) + ... + e (x);

The probability P =1-(1-1/x) ^b, which is selected in B-time, is only selected and not selected in each of the numbers in B.

So the expectation of each number is also i* (1-1/x) ^b)

Get the sum expectation.

#include <cstdio> #include <cstring> #include <algorithm>using namespace std;d ouble e[110000];d ouble F (Double X,int d) {    double ans = 1.0;    if (d = = 0) return 1.0;    ans = f (x,d/2);    Ans *= ans;    if (d%2)        ans *= x;    return ans;} int main () {    int t, TT, X, D, I;    scanf ("%d", &t);    for (TT = 1; TT <= t; tt++)    {        scanf ("%d%d", &x, &d);        for (i = 1; i <= x; i++)            e[i] = 1.0-f (1.0-1.0/x,d);        for (e[0] = 0, i = 1; i <= x; i++)            e[0] + = e[i]*i;        printf ("Case #%d:%.3lf\n", TT, E[0]);    }    return 0;}

Combination method

And all the results may appear and divide by the total kind.

#include <cstdio> #include <cstring> #include <algorithm>using namespace std;d ouble c[10]; int a[10][    int main () {int T, TT, X, B, N, I, J;    Double ans, k;    A[0][0] = 1;        for (i = 1; I <= 5; i++) {a[i][1] = 1;    A[i][i] = a[i-1][i-1] * i;    } a[3][2] = 6; A[4][2] = 14;    A[4][3] = 36; A[5][2] = 30; A[5][3] = 150;    A[5][4] = 240;    scanf ("%d", &t);        for (TT = 1; TT <= t; tt++) {ans = 0;        scanf ("%d%d", &x, &b);        n = min (x,b);        C[0] = 1; for (i = 1; I <= n-1; i++) {for (j = 1, k = 1; J <= I; j + +) {k *= (            (x-1)-j+1) *1.0/j;            } if (I >= 2) c[i] = k/(x*1.0*x);        else c[i] = k; } for (i = 1; I <= n; i++) {//printf ("C--%lf%lf\n", C[i-1], ((x+1) *1.0*x/2.0) * A[b][i])            ; K = c[i-1] * 1.0 * (x+1) *1.0*x/2.0) * A[b][i];            printf ("%lf\n", K);            int m = b;            if (I >= 3) m = b-2;            for (j = 1; j <= M; j + +) K/= (x*1.0);            printf ("%lf\n", K);        Ans + = k;    } printf ("Case #%d:%.3lf\n", tt, ANS); } return 0;}

Hdu5159--bc--card (probability notation, and combinatorial notation)