hdu4407 Sum 容斥原理

XXX is puzzled with the question below:

1, 2, 3, ..., n (1<=n<=400000) are placed in a line. There are m (1<=m<=1000) operations of two kinds.

Operation 1: among the x-th number to the y-th number (inclusive), get the sum of the numbers which are co-prime with p( 1 <=p <= 400000).
Operation 2: change the x-th number to c( 1 <=c <= 400000).

For each operation, XXX will spend a lot of time to treat it. So he wants to ask you to help him.

容斥原理

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

 const int maxn=4e5+;

 inline int gcd(int a,int b){return b?gcd(b,a%b):a;}

 int ori[],cha[];
 int pnum[],num;

 int main(){
     int T;
     scanf("%d",&T);
     while(T--){
         int n,m;
         scanf("%d%d",&n,&m);
         int cnt=;
         while(m--){
             int f;
             scanf("%d",&f);
             if(f==){
                 int x,y,p;
                 scanf("%d%d%d",&x,&y,&p);
                 int tmp=p;
                 num=;
                 for(int i=;i*i<=tmp;++i){
                     if(!(tmp%i)){
                         pnum[++num]=i;
                         tmp/=i;
                         while(!(tmp%i))tmp/=i;
                     }
                 }
                 if(tmp>)pnum[++num]=tmp;
                 ll ans=;
                 for(int i=;i<(<<num);++i){
                     int bit=;
                     ll mul=;
                     for(int j=;j<=num;++j){
                         if(i&(<<(j-))){
                             bit++;
                             mul*=pnum[j];
                         }
                     }
                     ll tmp=y/mul-(x-)/mul;
                     ll l=((x-)/mul+)*mul,r=y/mul*mul;
                     tmp=(l+r)*tmp/;
                     if(bit%)ans+=tmp;
                     else ans-=tmp;
                 }
                 ans=(x+y)*(ll)(y-x+)/-ans;
                 for(int i=;i<=cnt;++i){
                     if(ori[i]>=x&&ori[i]<=y){
                         int gcd1=gcd(ori[i],p),gcd2=gcd(cha[i],p);
                         if(gcd1==&&gcd2>)ans-=ori[i];
                         else if(gcd1>&&gcd2==)ans+=cha[i];
                         else if(gcd1==&&gcd2==)ans=ans-ori[i]+cha[i];
                     }
                 }
                 printf("%lld\n",ans);
             }
             else{
                 int x,c;
                 scanf("%d%d",&x,&c);
                 bool f=;
                 for(int i=;i<=cnt;++i){
                     if(ori[i]==x){
                         cha[i]=c;
                         f=;
                         break;
                     }
                 }
                 if(f){
                     ++cnt;
                     ori[cnt]=x;
                     cha[cnt]=c;
                 }
             }
         }
     }
     return ;
 }