hdu1695 GCD 容斥原理

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

 #include<stdio.h>
 #include<string.h>
 #include<algorithm>
 #include<math.h>
 using namespace std;
 typedef long long ll;

 const int maxn=2e5+;

 int pnum[maxn][],num[maxn];

 void init(){
     memset(pnum,,sizeof(pnum));
     memset(num,,sizeof(num));
     for(int i=;i<=;++i){
         if(!num[i]){
             pnum[i][++num[i]]=i;
             for(int j=;i*j<=;++j){
                 pnum[i*j][++num[i*j]]=i;
             }
         }
     }
 }

 int main(){
     init();
     int T;
     scanf("%d",&T);
     for(int q=;q<=T;++q){
         int a,b,c,d,k;
         scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);
         if(!k){
             printf("Case %d: 0\n",q);
             continue;
         }
         a=b/k;b=d/k;
         if(a>b){c=a;a=b;b=c;}
         ll ans=;
         if(a>=)ans+=b;
         for(int p=;p<=a;++p){
             ll res=;
             for(int i=;i<(<<num[p]);++i){
                 int bit=;
                 ll mul=;
                 for(int j=;j<=num[p];++j){
                     if(i&(<<(j-))){
                         ++bit;
                         mul*=pnum[p][j];
                     }
                 }
                 if(bit%)res+=(b/mul-(p-)/mul);
                 else res-=(b/mul-(p-)/mul);
             }
             ans+=b-(p-)-res;
         }
         printf("Case %d: %lld\n",q,ans);
     }
     return ;
 }