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HDU 4599 Dice

2024-07-17 02:26:10   221人阅读

Dice

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 211    Accepted Submission(s): 127


Problem Description
Given a normal dice (with 1, 2, 3, 4, 5, 6 on each face), we define: 
F(N) to be the expected number of tosses until we have a number facing up for N consecutive times.
H(N) to be the expected number of tosses until we have the number ‘1‘ facing up for N consecutive times.
G(M) to be the expected number of tosses until we have the number ‘1‘ facing up for M times.
Given N, you are supposed to calculate the minimal M1 that G (M1) >= F (N) and the minimal M2 that G(M2)>=H(N)
 

Input
The input contains multiple cases. 
Each case has a positive integer N in a separated line. (1<=N<=1000000000) 
The input is terminated by a line containing a single 0.
 

Output
For each case, output the minimal M1 and M2 as required in a single line, separated by a single space. 
Since the answer could be very large, you should output the answer mod 2011 instead.
 

Sample Input
1
2
0
 

Sample Output
1 1 
2 7
 

Source
2013 ACM-ICPC吉林通化全国邀请赛——题目重现
 

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设dp[i]为已经连续掷出i次相同点数,要到达目标状态所需的平均期望
则：dp[i] = (1/6)*(dp[i+1]+1) + (5/6)*(dp[1]+1)-----------------(1)
{ dp[0]是我们最终要求的值,dp[n] = 0 }


其中，下次掷出相同点数的概率为(1/6),再加上本身掷的一次,因此为(1/6)*(dp[i+1]+1)
如果下次掷的点数不同则相当于从1开始掷,因此为(5/6)*(dp[1]+1)
化简上式(1)可得：
dp[i] = (1/6)*(dp[i+1]) + (5/6)*(dp[1]) + 1------------------------(2)
dp[i+1] = (1/6)*(dp[i+2]) + (5/6)*(dp[1]) + 1---------------------(3)


两式相减(3) - (2)可得：
dp[i+1] - dp[i] = (1/6)*(dp[i+2] - dp[i+1])------------------------(4)


令a[i] = dp[i] - dp[i+1]-----------------------------------------------(5)
则a[i+1] = dp[i+2] - dp[i+1]
则(4)式变为：a[i+1] = 6*a[i]---------------------------------------(6)
由(2)式可知：dp[0] - dp[1] = 1,则a[0] = 1


因此a[i] = (1/6)^i------------------------------------------------------(7)


dp[n-1] - dp[n] = 6^(n-1)
dp[n-2] - dp[n-1] = 6^(n-2)
...
dp[0] - dp[1] = 1
则dp[0] = dp[n] + 6^0 + 6^1 + 6^2 + ... + 6^(n-2) + 6^(n-1)-----(8)
由于dp[n] = 0,则最终我们要求的结果dp[0]：
dp[0] = (6^n - 1)/5----------------------------------------------------(9)
即F[n] = (6^n - 1)/5--------------------------------------------------(10)
H[n] = 6*F[n] = 6*(6^n - 1)/5--------------------------------------(11)
G[m] = 6*m-----------------------------------------------------------(12)
由于题目要求使得G[m1]>=F[n] && G[m2]>=H[n]的最小整数值
则 m1 >= (6^n - 1)/30, m2>=(6^n - 1)/5
由于6^n尾数为6,因此要使得m1、m2为整数,则
m1 >= (6^n + 24)/30, m2>=(6^n - 1)/5
即求出在n的情况下上述两个表达式的值即可,
分母的情况用模n情况下的a的逆元来解决。

#include<stdio.h>
#include<iostream>
using namespace std;
typedef long long LL;
const LL MOD = 2011;
LL quick_mod(LL a,LL b,LL m)
{
    LL ans = 1;
    while(b)
    {
        if(b&1)
        {
            ans = (ans*a)%m;
            b--;
        }
        b/=2;
        a = a*a%m;
    }
    return ans;
}
int main()
{
    //F(n) = (6^n-1)/5-----H(n) = 6*F(n)-----G(m) = 6*m;
    //M1 = (6^n+24)/30-----M2 = (6^n-1)/5
    LL N;
    while(scanf("%I64d",&N) && N)
    {
        LL hehe,ans1,ans2;
        hehe = quick_mod(6,N,MOD);
        ans1 = ((hehe+24+MOD)%MOD*1944)%MOD;
        ans2 = ((hehe-1+MOD)%MOD*1609)%MOD;
        printf("%I64d %I64d\n",ans1,ans2);
    }
    return 0;
}



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