HDU 4407 Sum

分手后的思念是犯贱 2022-08-05 08:26 123阅读 0赞

这题一开始以为是什么数据结构，但是确实没想到什么东西维护和一个变化的互质的东西。但是如果是容斥就可以做，对于无修改的数列，和p互质的数的累加。  
但是因为各种操作数最大为1000，所以可以暴力维护修改的情况。最多不超过500∗500个查询。

// whn6325689
    // Mr.Phoebe
    // http://blog.csdn.net/u013007900
    #include <algorithm>
    #include <iostream>
    #include <iomanip>
    #include <cstring>
    #include <climits>
    #include <complex>
    #include <fstream>
    #include <cassert>
    #include <cstdio>
    #include <bitset>
    #include <vector>
    #include <deque>
    #include <queue>
    #include <stack>
    #include <ctime>
    #include <set>
    #include <map>
    #include <cmath>
    #include <functional>
    #include <numeric>
    #pragma comment(linker, "/STACK:1024000000,1024000000")
    
    
    using namespace std;
    #define eps 1e-9
    #define PI acos(-1.0)
    #define INF 0x3f3f3f3f
    #define LLINF 1LL<<50
    #define speed std::ios::sync_with_stdio(false);
    
    typedef long long ll;
    typedef unsigned long long ull;
    typedef long double ld;
    typedef pair<ll, ll> pll;
    typedef complex<ld> point;
    typedef pair<int, int> pii;
    typedef pair<pii, int> piii;
    typedef vector<int> vi;
    
    #define CLR(x,y) memset(x,y,sizeof(x))
    #define CPY(x,y) memcpy(x,y,sizeof(x))
    #define clr(a,x,size) memset(a,x,sizeof(a[0])*(size))
    #define cpy(a,x,size) memcpy(a,x,sizeof(a[0])*(size))
    #define debug(a) cout << #a" = " << (a) << endl;
    #define debugarry(a, n) for (int i = 0; i < (n); i++) { cout << #a"[" << i << "] = " << (a)[i] << endl; }
    
    #define mp(x,y) make_pair(x,y)
    #define pb(x) push_back(x)
    #define lowbit(x) (x&(-x))
    
    #define MID(x,y) (x+((y-x)>>1))
    #define ls (idx<<1)
    #define rs (idx<<1|1)
    #define lson ls,l,mid
    #define rson rs,mid+1,r
    
    template<class T>
    inline bool read(T &n)
    {
        T x = 0, tmp = 1;
        char c = getchar();
        while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();
        if(c == EOF) return false;
        if(c == '-') c = getchar(), tmp = -1;
        while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();
        n = x*tmp;
        return true;
    }
    template <class T>
    inline void write(T n)
    {
        if(n < 0)
        {
            putchar('-');
            n = -n;
        }
        int len = 0,data[20];
        while(n)
        {
            data[len++] = n%10;
            n /= 10;
        }
        if(!len) data[len++] = 0;
        while(len--) putchar(data[len]+48);
    }
    //-----------------------------------
    
    const int MAXN=400010;
    int s[MAXN][10],num[MAXN], a[10];
    ll ans;
    map<int,int> ma;
    
    int cal(int x,int p)
    {
        if(__gcd(x,p)>1)    return 0;
        return x;
    }
    
    void init()
    {
        for(int i=2; i<MAXN; i++)
            if(num[i]==0)
                for(int j=i; j<MAXN; j+=i)
                    s[j][++num[j]]=i;
    }
    
    int main()
    {
    
        int T,n,q;
        init();
        read(T);
        while(T--)
        {
            read(n),read(q);
            ma.clear();
            while(q--)
            {
                ll x, y;
                int op,p;
                read(op),read(x),read(y);
                if(op==1)
                {
                    read(p);
                    ans = (x+y)*(y-x+1)/2;
                    for (int i = 1; i <= num[p]; i++)
                    {
                        memset(a, 0, sizeof(a));
                        for (int j = num[p]; j >= num[p]-i+1; j--) a[j] = 1;
                        ll cnt = 0;
                        int tmp;
                        do
                        {
                            tmp = 1;
                            for (int k = 1; k <= num[p]; k++)
                            if (a[k] == 1) tmp = tmp*s[p][k];
                            cnt = y/tmp-(x-1)/tmp;
                            ll fir = tmp + ((x-1)/tmp)*tmp, las = tmp + (y/tmp-1)*tmp;
                            ll sum = (fir+las)*cnt/2;
                            if (i&1) ans -= sum;
                            else ans += sum;
                        } while (next_permutation(a+1, a+num[p]+1));
                    }
                    map<int,int>::iterator it,be=ma.lower_bound(x),en=ma.upper_bound(y);
                    for(it=be; it!=en; it++)
                        ans+=cal(it->second,p)-cal(it->first,p);
                    printf("%lld\n",ans);
                }
                else
                    ma[x]=y;
            }
        }
        return 0;
    }