Dice Possibility

Dice Possibility

时间限制:10000ms

单点时限:1000ms

内存限制:256MB

描述

What is possibility of rolling N dice and the sum of the numbers equals to M?

输入

Two integers N and M. (1 ≤ N ≤ 100, 1 ≤ M ≤ 600)

输出

Output the possibility in percentage with 2 decimal places.

样例输入

2 10

样例输出

8.33
分析：依次枚举骰子；
代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#include <ext/rope>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define vi vector<int>
#define pii pair<int,int>
#define mod 1000000007
#define inf 0x3f3f3f3f
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
const int maxn=1e3+10;
const int dis[][2]={0,1,-1,0,0,-1,1,0};
using namespace std;
using namespace __gnu_cxx;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
int n,m;
double dp[maxn][maxn];
int main()
{
    int i,j,k,t;
    scanf("%d%d",&n,&m);
    rep(i,1,6)dp[1][i]=1.0/6;
    rep(i,2,n)
    {
        rep(j,2,m)rep(k,1,6)
            if(j-k>0)dp[i][j]+=dp[i-1][j-k]/6;
    }
    printf("%.2f
",dp[n][m]*100);
    //system("pause");
    return 0;
}