Given 5 integers: a, b, c, d, k, you're to find x in a...b, y in c...d that GCD(x, y) = k. GCD(x, y) means the greatest common divisor of x and y. Since the number of choices may be very large, you're only required to output the total number of different number pairs.
Please notice that, (x=5, y=7) and (x=7, y=5) are considered to be the same.
Yoiu can assume that a = c = 1 in all test cases.
容斥原理的裸题
1 #include<stdio.h>
2 #include<string.h>
3 #include<algorithm>
4 #include<math.h>
5 using namespace std;
6 typedef long long ll;
7
8 const int maxn=2e5+5;
9
10 int pnum[maxn][30],num[maxn];
11
12 void init(){
13 memset(pnum,0,sizeof(pnum));
14 memset(num,0,sizeof(num));
15 for(int i=2;i<=100100;++i){
16 if(!num[i]){
17 pnum[i][++num[i]]=i;
18 for(int j=2;i*j<=100100;++j){
19 pnum[i*j][++num[i*j]]=i;
20 }
21 }
22 }
23 }
24
25 int main(){
26 init();
27 int T;
28 scanf("%d",&T);
29 for(int q=1;q<=T;++q){
30 int a,b,c,d,k;
31 scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);
32 if(!k){
33 printf("Case %d: 0\n",q);
34 continue;
35 }
36 a=b/k;b=d/k;
37 if(a>b){c=a;a=b;b=c;}
38 ll ans=0;
39 if(a>=1)ans+=b;
40 for(int p=2;p<=a;++p){
41 ll res=0;
42 for(int i=1;i<(1<<num[p]);++i){
43 int bit=0;
44 ll mul=1;
45 for(int j=1;j<=num[p];++j){
46 if(i&(1<<(j-1))){
47 ++bit;
48 mul*=pnum[p][j];
49 }
50 }
51 if(bit%2)res+=(b/mul-(p-1)/mul);
52 else res-=(b/mul-(p-1)/mul);
53 }
54 ans+=b-(p-1)-res;
55 }
56 printf("Case %d: %lld\n",q,ans);
57 }
58 return 0;
59 }
