GCJ 2008 APAC local onsites C Millionaire

/*非常值得重新重做的概率题，化连续为离散，分为(2^M +1)个区间，double型的赌注，则用该区间内的int型来代替，反正效果一样(最终得到的概率是相等的)等到要输出prv[i]时，再去找在dp数组中，该本金对应的是(2^M+1)个区间中的哪个区间*/#include <cstdio>#include <iostream>#include <algorithm>#include <cstring>#include <iomanip>using namespace std;typedef long long ll;const int MAX_M = 15;int M, X;double P;double dp[2][(1 << MAX_M) + 1];void solve(){int n = 1 << M;double *prv = dp[0], *nxt = dp[1];memset(prv, 0, sizeof (double ) * (n + 1));prv[n] = 1.0;for (int r = 0; r < M; r++) //枚举轮{for (int i = n; i >= 0; i--){int jub = min(i, n - i);double t = 0.0;for (int j = 0; j <= jub; j++){t = max(t, P * prv[i + j] + (1 - P) * prv[i - j]);}nxt[i] = t;} swap(prv, nxt);}int i = (ll) X * n / (1e6);//cout << fixed << setprecision(6) << prv[i] << endl;printf("%.6f\n", prv[i]);}int main(){freopen("E:\\c2.txt", "r", stdin);freopen("E:\\out2.txt", "w", stdout);int k;//cin >> k;scanf("%d",&k);for (int kase = 1; kase <= k; kase++){//cin >> M >> P >> X;scanf("%d%lf%d", &M, &P, &X);printf("Case #%d: ", kase);//cout << "Case #" << kase << ": ";solve(); }fclose(stdin);fclose(stdout);return 0;}