hdu 4790 Just Random （2013成都J题） 数学思路题 容斥

 1 #include <stdio.h>
 2 #include <algorithm>
 3 #include <string.h>
 4 #include<cmath>
 5 #define LL long long
 6 using namespace std;
 7 int T;
 8 LL a,b,c,d,p,m;
 9 
10 LL gcd(LL a, LL b) {
11     return b ? gcd(b, a % b) : a;
12 }
13 
14 LL fun(LL x,LL y) {//表示0到x区间，0到y区间的组合对数
15     LL ret;
16     LL ra,rb;
17     ra=x%p,rb=y%p;
18     ret=(x/p)*(y/p)*p;
19     ret+=(ra+1)*(y/p)+(rb+1)*(x/p);
20     if(ra>m) {
21         ret+=min(m+1,rb+1);
22         LL tmp=(m+p-ra)%p;
23         if(tmp<=rb) {
24             ret+=rb-tmp+1;
25         }
26     } else {
27         LL tmp=(m+p-ra)%p;
28         if(tmp<=rb) {
29             ret+=min(m-tmp+1,rb-tmp+1);
30         }
31     }
32     return ret;
33 }
34 int main() {
35     scanf("%d",&T);
36     int cas=0;
37     while(T--) {
38         cas++;
39         scanf("%I64d%I64d%I64d%I64d%I64d%I64d",&a,&b,&c,&d,&p,&m);
40         LL ans=fun(b,d)-fun(a-1,d)-fun(b,c-1)+fun(a-1,c-1);//容斥原理求和
41         LL tot=(b-a+1)*(d-c+1);
42         if(ans<0)
43             ans=0;
44 //        printf("ans:%I64d tot:%I64d\n",ans,tot);
45         LL l=gcd(ans,tot);
46         ans/=l,tot/=l;
47         printf("Case #%d: %I64d/%I64d\n", cas, ans, tot);
48     }
49     return 0;
50 }