HDU 5634 线段树

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5634

题意：给定一个长度为n的序列，有m次操作。操作有3种：

1 l,r :区间[l,r]的值变成phi[val[i]](l<=i<=r; phi是欧拉值)

2 l,r,x:区间[l,r]的值变成x

3 l,r:求区间[l,r]的和

思路：操作2和3就是传统的简单线段树，操作2对应区间覆盖，操作3对应区间求和，重点在于操作1，由于一个数经过不超过log次求phi后会变成1，所以可以在线段树是用一个same标记，如果整个区间的数都相同则操作1就转换成操作2的区间覆盖了。如果操作的区间[l,r]已经包含住当前递归的子树区间但是子树的same标记为假则继续递归到子树的same标记为真为止，最多递归到叶子结点。 

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<queue>
#include<vector>
#include<time.h>
#include<cmath>
using namespace std;
typedef long long int LL;
#define L(k)(k<<1)
#define R(k)(k<<1|1)
const LL INF = 9223372036854775807;
const int MAXN = 3e5 + 24;
const int MAXX = 1e7 + 24;
struct Node{
    int l, r, val;
    LL sum;
    bool same;
    Node(int _l = 0, int _r = 0, int _val = 0, LL _sum = 0, bool _same = false){
        l = _l; r = _r; sum = _sum; same = _same; val = _val;
    }
}Seg[MAXN * 4];
int val[MAXN], Phi[MAXX];
void pushUp(int k){
    Seg[k].sum = Seg[L(k)].sum + Seg[R(k)].sum;
    if (Seg[L(k)].val == Seg[R(k)].val&&Seg[L(k)].same&&Seg[R(k)].same&&Seg[L(k)].val != -1){
        Seg[k].same = true;
        Seg[k].val = Seg[L(k)].val;
    }
    else{
        Seg[k].same = false;
        Seg[k].val = -1;
    }
}
void pushDown(int k){
    if (Seg[k].same){
        Seg[L(k)].same = Seg[R(k)].same = true;
        Seg[L(k)].val = Seg[R(k)].val = Seg[k].val;
        Seg[L(k)].sum = 1LL * (Seg[L(k)].r - Seg[L(k)].l + 1)*Seg[L(k)].val;
        Seg[R(k)].sum = 1LL * (Seg[R(k)].r - Seg[R(k)].l + 1)*Seg[R(k)].val;
    }
}
void Build(int st, int ed, int k){
    Seg[k].l = st; Seg[k].r = ed; Seg[k].same = false;  Seg[k].val = -1;
    if (st == ed){
        Seg[k].val = val[st];
        Seg[k].sum = val[st];
        Seg[k].same = true;
        return;
    }
    int mid = (st + ed) >> 1;
    Build(st, mid, L(k)); Build(mid + 1, ed, R(k));
    pushUp(k);
}
void Change(int st, int ed, int val, int k){
    if (Seg[k].l == st&&Seg[k].r == ed){
        Seg[k].same = true;
        Seg[k].val = val;
        Seg[k].sum = 1LL * (ed - st + 1)*val;
        return;
    }
    pushDown(k);
    if (Seg[L(k)].r >= ed){
        Change(st, ed, val, L(k));
    }
    else if (Seg[R(k)].l <= st){
        Change(st, ed, val, R(k));
    }
    else{
        Change(st, Seg[L(k)].r, val, L(k));
        Change(Seg[R(k)].l, ed, val, R(k));
    }
    pushUp(k);
}
void Modify(int st, int ed, int k){
    if (Seg[k].l == st&&Seg[k].r == ed&&Seg[k].same){
        Seg[k].val = Phi[Seg[k].val];
        Seg[k].sum = 1LL * (ed - st + 1)*Seg[k].val;
        return;
    }
    pushDown(k);
    if (Seg[L(k)].r >= ed){
        Modify(st, ed, L(k));
    }
    else if (Seg[R(k)].l <= st){
        Modify(st, ed, R(k));
    }
    else{
        Modify(st, Seg[L(k)].r, L(k));
        Modify(Seg[R(k)].l, ed, R(k));
    }
    pushUp(k);
}
LL Query(int st, int ed, int k){
    if (Seg[k].l == st&&Seg[k].r == ed){
        return Seg[k].sum;
    }
    pushDown(k);
    if (Seg[L(k)].r >= ed){
        return Query(st, ed, L(k));
    }
    else if (Seg[R(k)].l <= st){
        return Query(st, ed, R(k));
    }
    else{
        return Query(st, Seg[L(k)].r, L(k)) + Query(Seg[R(k)].l, ed, R(k));
    }
    pushUp(k);
}
void GetPhi(){ //预处理欧拉值
    memset(Phi, 0, sizeof(Phi));
    Phi[1] = 1;
    for (LL i = 2; i < MAXX; i++){
        if (!Phi[i]){
            for (LL j = i; j < MAXX; j += i){
                if (!Phi[j]) Phi[j] = j;
                Phi[j] = Phi[j] / i*(i - 1);
            }
        }
    }
}
int main(){
//#ifdef kirito
//    freopen("in.txt", "r", stdin);
//    freopen("out.txt", "w", stdout);
//#endif
//    int start = clock();
    int n, t, m; GetPhi();
    scanf("%d", &t);
    while (t--){
        scanf("%d%d", &n, &m);
        for (int i = 1; i <= n; i++){
            scanf("%d", &val[i]);
        }
        Build(1, n, 1);
        for (int i = 1; i <= m; i++){
            int tpe, l, r, x;
            scanf("%d%d%d", &tpe, &l, &r);
            if (tpe == 1){ 
                Modify(l, r, 1);
            }
            else if (tpe == 2){
                scanf("%d", &x);
                Change(l, r, x, 1);
            }
            else{
                printf("%lld
", Query(l, r, 1));
            }
        }
    }
//#ifdef LOCAL_TIME
//    cout << "[Finished in " << clock() - start << " ms]" << endl;
//#endif
    return 0;
}