Light OJ 1064 - Throwing Dice

Posted 2020-07-17 GFY

题目链接：http://www.lightoj.com/volume_showproblem.php?problem=1064

题意：一次掷n个筛子，求点数之和大于m的概率。

思路：f[i][j]表示前i个筛子点数之和为j的概率，由此DP即可。不过这道题不用每次都乘以六分之一，因为n个筛子最后的分母是6^n，所以最后除以6^n即可。

 1 #include<algorithm>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<iostream>
 6 #define ll unsigned long long
 7 ll f[30][160],p[30];
 8 int n,m;
 9 ll gcd(ll a,ll b){
10     if (b==0) return a;
11     else return gcd(b,a%b);
12 }
13 int main(){
14     int T;
15     scanf("%d",&T);
16     p[0]=1;
17     for (int i=1;i<=24;i++) p[i]=p[i-1]*6;
18     for (int i=1;i<=6;i++) f[1][i]=1;
19     for (int i=2;i<=24;i++)
20      for (int j=i-1;j<=(i-1)*6;j++)
21       for (int k=1;k<=6;k++)
22        f[i][j+k]+=f[i-1][j];
23     for (int i=1;i<=24;i++)
24      for (int j=i*6;j>=0;j--)
25       f[i][j]+=f[i][j+1];
26     for (int Tcase=1;Tcase<=T;Tcase++){
27         scanf("%d%d",&n,&m);
28         ll ans1=f[n][m];
29         ll ans2=p[n];
30         ll t=gcd(ans1,ans2);
31         ans1/=t;ans2/=t;
32         if (ans2==1)
33         printf("Case %d: %lld\n",Tcase,ans1);
34         else
35         printf("Case %d: %lld/%lld\n",Tcase,ans1,ans2);
36     }
37 }