E. Kleofáš and the n-thlon time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard output Kleofáš is participating in an n-thlon - a tournament consisting of n different competitions in n different disciplines (numbered 1 through n). There are m participants in the n-thlon and each of them participates in all competitions.
In each of these n competitions, the participants are given ranks from 1 to m in such a way that no two participants are given the same rank - in other Words, the ranks in each competition form a permutation of numbers from 1 to m. The score of a participant in a competition is equal to his/her rank in it.
The overall score of each participant is computed as the sum of that participant’s scores in all competitions.
The overall rank of each participant is equal to 1 + k, where k is the number of participants with strictly smaller overall score.
The n-thlon is over now, but the results haven’t been published yet. Kleofáš still remembers his ranks in each particular competition; however, he doesn’t remember anything about how well the other participants did. Therefore, Kleofáš would like to know his expected overall rank.
All competitors are equally good at each discipline, so all rankings (permutations of ranks of everyone except Kleofáš) in each competition are equiPRobable.
Input The first line of the input contains two space-separated integers n (1 ≤ n ≤ 100) and m (1 ≤ m ≤ 1000) — the number of competitions and the number of participants respectively.
Then, n lines follow. The i-th of them contains one integer xi (1 ≤ xi ≤ m) — the rank of Kleofáš in the i-th competition.
Output Output a single real number – the expected overall rank of Kleofáš. Your answer will be considered correct if its relative or absolute error doesn’t exceed 10 - 9.
Namely: let’s assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if .
Examples input 4 10 2 1 2 1 output 1.0000000000000000 input 5 5 1 2 3 4 5 output 2.7500000000000000 input 3 6 2 4 2 output 1.6799999999999999 Note In the first sample, Kleofáš has overall score 6. Nobody else can have overall score less than 6 (but it’s possible for one other person to have overall score 6 as well), so his overall rank must be 1.
这个题的关键点在于他没法确切的求出每个人期望… 看上去一团糟…. 第一次做这种题… 哇太弱了还是看了一晚上题解才学会的 除了自己以外其他人都是一样的 那么只要求一个人的概率就够了 那么你当前场次的分数可以从上一场的当年分数-1到-m里去找 每一个都是m-1分之一的概率转移过来 要注意的是自己的那个不能转移过来 初始设定为m-1 是因为那个值要处以m-1 实际上每一个第一场的概率一定都是1
#include<iostream>#include<algorithm>#include<cmath>#include<cstdio>using namespace std;int tu[101];double he[101][100101], dp[101][100101];int main(){ int n, m,ss=0; cin >> n >> m; if (m == 1) { printf("1.000000000000"); return 0; } dp[0][0] = m - 1; for (int a = 1;a <= n;a++)cin >> tu[a],ss+=tu[a]; for (int a = 1;a <= m*n+1;a++)he[0][a] =m-1; for (int a = 1;a <= n;a++) { he[a][0] = he[a][1] = 0; for (int b = 1;b <=n*m;b++) { int zuo = max(0, b - m), you = b; dp[a][b] += (he[a - 1][you] - he[a - 1][zuo])*1.0 / (m - 1); if (b - tu[a] >= 0)dp[a][b] -= dp[a - 1][b - tu[a]] / (m - 1); he[a][b+1] = he[a][b] + dp[a][b]; } } printf("%.9f", he[n][ss] + 1);}新闻热点
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