Throwing Dice LightOJ - 1064 || （勉强能用的）分数类

 1 #include<cstdio>
 2 typedef long long LL;
 3 LL gcd(LL a,LL b)
 4 {
 5     LL t;
 6     while(b!=0)
 7     {
 8         t=a;
 9         a=b;
10         b=t%b;
11     }
12     return a;
13 }
14 struct X
15 {
16     LL a,b;
17     X()
18     {
19         a=0;b=1;
20     }
21     X(int x,int y)
22     {
23         a=x;b=y;
24     }
25     X operator+(const X& c)
26     {
27         X ans;
28         LL tmp=gcd(b,c.b);
29         ans.b=b/tmp*c.b;
30         ans.a=ans.b/b*a+ans.b/c.b*c.a;
31         tmp=gcd(ans.a,ans.b);
32         ans.a/=tmp;
33         ans.b/=tmp;
34         return ans;
35     }
36     X operator*(const X& c)
37     {
38         LL tmp1=gcd(a,c.b),tmp2=gcd(b,c.a);
39         X ans;
40         ans.a=a/tmp1*c.a/tmp2;
41         ans.b=b/tmp2*c.b/tmp1;
42         return ans;
43     }
44     void print()
45     {
46         if(b==1)
47             printf("%lld",a);
48         else
49             printf("%lld/%lld",a,b);
50     }
51 }ans[26][151],ans2[26][151];
52 LL T,TT,n,x;
53 int main()
54 {
55     int i,j,k;
56     for(i=1;i<=6;i++)
57         ans[1][i]=(X){1,6};
58     for(i=2;i<=25;i++)
59         for(j=i;j<=6*i;j++)
60             for(k=1;k<=6;k++)
61                 if(j-k>0)
62                     ans[i][j]=ans[i][j]+ans[i-1][j-k]*ans[1][1];//ans[1][1]就是1/6
63     for(i=1;i<=25;i++)
64     {
65         ans2[i][6*i]=ans[i][6*i];
66         for(j=6*i-1;j>=0;j--)
67             ans2[i][j]=ans2[i][j+1]+ans[i][j];
68     }
69     scanf("%lld",&T);
70     for(TT=1;TT<=T;TT++)
71     {
72         scanf("%lld%lld",&n,&x);
73         printf("Case %lld: ",TT);
74         ans2[n][x].print();
75         puts("");
76     }
77     return 0;
78 }