银川区域赛现场赛 Pot!!【线段树】

给定两个操作：

MULTIPLY L R x  区间里都乘以一个数x

MAX L R : 计算区间内一个2,3,5,7个数最大值。

思路：维护4个最大值。蓝瘦。

 /** 有 n 个数和 5 种操作
 add a b c：把区间[a,b]内的所有数都增加 c
 set a b c：把区间[a,b]内的所有数都设为 c
 sum a b：查询区间[a,b]的区间和
 max a b：查询区间[a,b]的最大值
 min a b：查询区间[a,b]的最小值
 */
 #include <bits/stdc++.h>
 using namespace std;
 const int maxn = 2e5 + ;
 const long long INF = 1LL << ;
 int Pri[]={  ,  ,  ,  };
 struct Segment_tree
 {
     struct Node
     {
         int l, r;
         long long sum, max, min, set_lazy, add_lazy;
     } tre[maxn << ];
     long long arr[maxn*];//需要输入的数组
     inline void push_up(int rt)
     {
         if(tre[rt].l == tre[rt].r)
         {
             return ;
         }
         tre[rt].sum = tre[rt<<].sum + tre[rt<<|].sum;
         tre[rt].max = max(tre[rt<<].max, tre[rt<<|].max);
         tre[rt].min = min(tre[rt<<].min, tre[rt<<|].min);
     }
     inline void push_down(int rt)
     {
         if(tre[rt].set_lazy) {
             ///if set_lazy add_lazy = 0
             tre[rt<<].set_lazy = tre[rt].set_lazy;
             tre[rt<<].sum = (tre[rt<<].r - tre[rt<<].l + ) * tre[rt].set_lazy;
             tre[rt<<].max = tre[rt].set_lazy;
             tre[rt<<].min = tre[rt].set_lazy;
             tre[rt<<|].set_lazy = tre[rt].set_lazy;
             tre[rt<<|].sum = (tre[rt<<|].r - tre[rt<<|].l + ) * tre[rt].set_lazy;
             tre[rt<<|].max = tre[rt].set_lazy;
             tre[rt<<|].min = tre[rt].set_lazy;
             tre[rt].add_lazy = ;
             tre[rt<<].add_lazy = tre[rt<<|].add_lazy = ;
             tre[rt].set_lazy = ;
             return ;
         }
         if(tre[rt].add_lazy)
         {
             tre[rt<<].add_lazy += tre[rt].add_lazy;
             tre[rt<<].sum += (tre[rt<<].r - tre[rt<<].l + ) * tre[rt].add_lazy;
             tre[rt<<].max += tre[rt].add_lazy;
             tre[rt<<].min += tre[rt].add_lazy;
             tre[rt<<|].add_lazy += tre[rt].add_lazy;
             tre[rt<<|].sum += (tre[rt<<|].r - tre[rt<<|].l + ) *
                                 tre[rt].add_lazy;
             tre[rt<<|].max += tre[rt].add_lazy;
             tre[rt<<|].min += tre[rt].add_lazy;
             tre[rt].add_lazy = ;
         }
     }
     void build(int rt,int l,int r)
     {
         tre[rt].l = l;
         tre[rt].r = r;
         tre[rt].set_lazy = ;
         tre[rt].add_lazy = ;
         if(l == r)
         {
             tre[rt].sum = tre[rt].max = tre[rt].min = arr[l];
             return ;
         }
         int mid = (l + r) >> ;
         build(rt<<,l,mid);
         build(rt<<|,mid+,r);
         push_up(rt);
     }
     void update1(int rt,int l,int r,long long val)///add
     {
         push_down(rt);
         if(l == tre[rt].l && tre[rt].r == r)
         {
             tre[rt].add_lazy = val;
             tre[rt].sum += (tre[rt].r - tre[rt].l + ) * val;
             tre[rt].max += val;
             tre[rt].min += val;
             return ;
         }
         int mid = (tre[rt].l + tre[rt].r) >> ;
         if(r <= mid)
         {
             update1(rt<<,l,r,val);
         }
         else if(l > mid)
         {
             update1(rt<<|,l,r,val);
         }
         else
         {
             update1(rt<<,l,mid,val);
             update1(rt<<|,mid+,r,val);
         }
         push_up(rt);
     }
     void update2(int rt,int l,int r,long long val)///set
     {
         push_down(rt);
         if(l == tre[rt].l && tre[rt].r == r) {
             tre[rt].set_lazy = val;
             tre[rt].sum = (tre[rt].r - tre[rt].l + ) * val;
             tre[rt].max = val;
             tre[rt].min = val;
             tre[rt].add_lazy = ;
             return ;
         }
         int mid = (tre[rt].l + tre[rt].r) >> ;
         if(r <= mid) {
             update2(rt<<,l,r,val);
         } else if(l > mid) {
             update2(rt<<|,l,r,val);
         } else {
             update2(rt<<,l,mid,val);
             update2(rt<<|,mid+,r,val);
         }
         push_up(rt);
     }
     long long query1(int rt,int l,int r)///sum
     {
         push_down(rt);
         if(l == tre[rt].l && tre[rt].r == r) {
             return tre[rt].sum;
         }
         int mid = (tre[rt].l + tre[rt].r) >> ;
         if(r <= mid) {
             return query1(rt<<,l,r);
         } else if(l > mid) {
             return query1(rt<<|,l,r);
         } else {
             return query1(rt<<,l,mid) + query1(rt<<|,mid+,r);
         }
     }
     long long query2(int rt,int l,int r)///max
     {
         push_down(rt);
         if(l == tre[rt].l && tre[rt].r == r) {
             return tre[rt].max;
         }
         int mid = (tre[rt].l + tre[rt].r) >> ;
         if(r <= mid) {
             return query2(rt<<,l,r);
         } else if(l > mid) {
             return query2(rt<<|,l,r);
         } else {
             return max(query2(rt<<,l,mid), query2(rt<<|,mid+,r));
         }
     }
     long long query3(int rt,int l,int r)///min
     {
         push_down(rt);
         if(l == tre[rt].l && tre[rt].r == r) {
             return tre[rt].min;
         }
         int mid = (tre[rt].l + tre[rt].r) >> ;
         if(r <= mid) {
             return query3(rt<<,l,r);
         } else if(l > mid) {
             return query3(rt<<|,l,r);
         } else {
             return min(query3(rt<<,l,mid), query3(rt<<|,mid+,r));
         }
     }
 }st[];
 int main(){
     int n,m;
     scanf("%d%d",&n,&m);
     for(int i=;i<;i++)
         st[i].build(,,n);
     while(m--){
         char  Q[];
         scanf("%s",Q);
         int L,R;
         if(Q[]=='U'){
             int x;
             scanf("%d%d%d",&L,&R,&x);
             for(int i=;i<;i++){
                 if(x%Pri[i]==){
                     while(x%Pri[i]==){
                         st[i].update1(,L,R,);
                         x/=Pri[i];
                     }
                 }
             }
         }else{
             scanf("%d%d",&L,&R);
             long long int ans=;
             for(int i=;i<;i++){
                 ans=max(ans,st[i].query2(,L,R));
             }
             printf("ANSWER ");
             printf("%lld\n",ans);
         }
     }
     return ;
 }

 /*
 5 6
 MULTIPLY 3 5 2
 MULTIPLY 2 5 3

 MAX 1 5
 MULTIPLY 1 4 2
 MULTIPLY 2 5 5
 MAX 3 5

 */