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Codeforces Round #333 (Div. 2)E

2019-11-11 07:51:18
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E. Kleofá? and the n-thlon time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard output Kleofá? 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. Kleofá? still remembers his ranks in each particular competition; however, he doesn’t remember anything about how well the other participants did. Therefore, Kleofá? 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 Kleofá?) 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 Kleofá? in the i-th competition.

Output Output a single real number – the expected overall rank of Kleofá?. 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, Kleofá? 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.

這個(gè)題的關(guān)鍵點(diǎn)在于他沒法確切的求出每個(gè)人期望… 看上去一團(tuán)糟…. 第一次做這種題… 哇太弱了還是看了一晚上題解才學(xué)會的 除了自己以外其他人都是一樣的 那么只要求一個(gè)人的概率就夠了 那么你當(dāng)前場次的分?jǐn)?shù)可以從上一場的當(dāng)年分?jǐn)?shù)-1到-m里去找 每一個(gè)都是m-1分之一的概率轉(zhuǎn)移過來 要注意的是自己的那個(gè)不能轉(zhuǎn)移過來 初始設(shè)定為m-1 是因?yàn)槟莻€(gè)值要處以m-1 實(shí)際上每一個(gè)第一場的概率一定都是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);}
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