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http://acm.hit.edu.cn/hoj/problem/view?id=3152
Dice My Tags (Edit) Source : Time limit : 1 sec Memory limit : 128 M Submitted : 82, Accepted : 18 You have a dice with M faces, each face contains a distinct number. Your task is to calculate the expected number of tosses until a number facing up for consecutive N times. We assume when we tossing the dice, each face will occur randomly and uniformly. Input Each test cases contains only one line with two integers M, N. (1<=M, N<=100000000) Output For each test case, display a single line with the answer to the question above, the answer maybe very large, you need to MOD it with 1000000007. Sample Input 5 6 1 6 2 6 3 6 4 6 5 Sample Output 1 7 43 259 1555
题目说是求M面骰子投出N次相同点数所需要投的次数的期望值,看样例可以知道其实是求1+M+M^2+.......+M^(N-1),等比数列求和取模!
这个好像是某次校赛的题,当时我是直接等比数列慢慢加起来,没用等比数列求和公式,就得了。不过现在这题数据好像加强了…不能这样撸啦!而等比数列求和公式又有除法,不好搞取模。于是我也不会了。
然后我找到了一个叫kk303的人写的这个题解:
http://blog.csdn.net/kk303/article/details/9332513
虽然还是不太懂为什么不过得到了超碉的公式:
求等比为k的等比数列之和T[n]..当n为偶数..T[n] = T[n/2] + pow(k,n/2) * T[n/2]
n为奇数...T[n] = T[n/2] + pow(k,n/2) * T[n/2] + 等比数列第n个数的值
比如 1+2+4+8 = (1+2) + 4*(1+2)
1+2+4+8+16 = (1+2) + 4*(1+2) + 16
哇,终于A了,简直屁滚尿流
代码:
1 #include<stdio.h> 2 #define MO 1000000007 3 long long m,n; 4 long long powmod(long long a,long long b) 5 { 6 long long c=1; 7 while(b>0) 8 { 9 if(b%2!=0) 10 c=c*a%MO; 11 a=a*a%MO; 12 b=b/2; 13 } 14 return c; 15 } 16 17 long long T(long long n) 18 { 19 if(n<=1) return 1; 20 long long TN2=T(n/2); 21 if(n%2==0) 22 { 23 return (TN2 + powmod(m,n/2) * TN2)%MO; 24 } 25 else 26 { 27 return (TN2 + powmod(m,n/2) * TN2 + powmod(m,n-1))%MO; 28 } 29 } 30 31 int main() 32 { 33 long long ans,k; 34 long long i,j,t; 35 scanf("%lld",&t); 36 for(i=1; i<=t; i++) 37 { 38 scanf("%lld%lld",&m,&n); 39 if(m==1) ans=n; 40 else 41 { 42 ans=T(n); 43 } 44 printf("%lld\n",ans); 45 } 46 return 0; 47 }
hoj3152-Dice 等比数列求和取模,布布扣,bubuko.com
标签:style class blog code http tar
原文地址:http://www.cnblogs.com/yuiffy/p/3809176.html