Codeforces 601C Kleofáš and the n-thlon（dp）

来源：互联网发布：iphone手机主题软件编辑：程序博客网时间：2024/05/11 06:00

题目链接：Codeforces 601C Kleofáš and the n-thlon

解题思路

dp[i][j]表示到第i场比赛时，得分为j的人数期望。dp[i][j] = sum { dp[i-1][j-x] ~ dp[i-1][j-x] | 1 ≤ x ≤ m && x != rank[i] }

代码

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 2 * 1e5 + 5;int N, M, X[105];double dp[2][maxn];int main () {    scanf("%d%d", &N, &M);    for (int i = 1; i <= N; i++) scanf("%d", &X[i]);    if (M == 1)        printf("%.10lf\n", 1.0);    else {        int nw = 0, nx = 1, s = 0;        memset(dp[nx], 0, sizeof(dp[nx]));        dp[nx][0] = M - 1;        for (int i = 1; i <= N; i++) {            swap(nw, nx);            memset(dp[nx], 0, sizeof(dp[nx]));            for (int j = 0; j <= i * M; j++) {                double p = dp[nw][j] / (M-1);                dp[nx][j+1] += p;                dp[nx][j+M+1] -= p;                dp[nx][j + X[i]] -= p;                dp[nx][j + X[i] + 1] += p;            }            for (int j = 1; j <= (i + 1) * M; j++)                dp[nx][j] += dp[nx][j-1];            s += X[i];        }        double ans = 0;        for (int i = 0; i < s; i++) ans += dp[nx][i];        printf("%.10lf\n", ans + 1);    }    return 0;}

0 0