http://acm.hdu.edu.cn/showproblem.php?pid=4407

把修改和询问分成两部分解决

询问求区间内与p不互素的和,和求个数一样,用容斥原理解决,只不过做容斥的时候把每一段的个数改成每一段的和,这个求和的方式我一下写搓了,导致这道题坑了很久

修改用map记录,每次扫一遍即可,原数和c互素就减掉,修改完的数和c互素就加上去,logn*m^2的复杂度

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <map>
#include <cmath>

using namespace std;

typedef __int64 ll;

ll gcd(ll a,ll b){
    return b==0?a:gcd(b,a%b);
}

ll sum(ll l,ll r,ll k){
    ll num=r/k-(l-1)/k;
    ll front;
    if(l%k==0)front=l;
    else front=(k-l%k)+l;
    ll rear=r-r%k;
    return (front+rear)*num/2;
}

ll gao(ll l,ll r,ll n){
    vector <ll> v;
    for(ll i=2;i*i<=n;i++){
        if(n%i==0){
            v.push_back(i);
            while(n%i==0)n/=i;
        }
    }
    if(n>1)v.push_back(n);
    int m=v.size();
    ll res=0; 
    for(int i=1;i<(1<<m);i++){
        int cnt=0;
        ll val=1;
        for(int j=0;j<m;j++){
            if(i&(1<<j)){
                cnt++;
                val*=v[j];
            }
        }
        if(cnt&1)res+=sum(l,r,val);
        else res-=sum(l,r,val);
    }
    return (l+r)*(r-l+1)/2-res;
}

map <ll,ll> mp;
map <ll,ll>::iterator it;

ll gan(ll x,ll y,ll p){
    ll res=gao(x,y,p);
    for(it=mp.begin();it!=mp.end();it++){
        if(it->first>y || it->first<x)continue;
        if(gcd(it->first,p)==1)res-=it->first;
        if(gcd(it->second,p)==1)res+=it->second;
    }
    return res;
}

int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        int n,m;
        scanf("%d%d",&n,&m);
        mp.clear();
        while(m--){
            int op;
            scanf("%d",&op);
            ll x,y,p;
            if(op==1){
                scanf("%I64d%I64d%I64d",&x,&y,&p);
                if(x>y)swap(x,y);
                printf("%I64d\n",gan(x,y,p));
            }
            else{
                scanf("%I64d%I64d",&x,&p);
                mp[x]=p;
            }
        }
    }
    return 0;
}
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