UESTC 电子科大专题训练 数据结构 C

UESTC 1597t

题意：中文题

思路：将每个点展开为 x*lazy1+lazy2， lazy1标记乘了多少 lazy2标记加了多少，每次区间乘的时候即 sum*c=(x*lazy1+lazy2)*c=x*lazy1*c+lazy2*c，区间加的时候为sum+c=(x*lazy1+lazy2)+c=x*lazy1+lazy2+c，每次对lazy标记的时候乘要同时对lazy1和lazy2进行标记

AC代码：

#include "iostream"
#include "string.h"
#include "stack"
#include "queue"
#include "string"
#include "vector"
#include "set"
#include "map"
#include "algorithm"
#include "stdio.h"
#include "math.h"
#define ll long long
#define bug(x) cout<<x<<" "<<"UUUUU"<<endl;
#define mem(a) memset(a,0,sizeof(a))
#define mp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
#define lrt (root*2)
#define rrt (root*2+1)
#define len (r-l+1)
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const long long INF = 1e18+1LL;
const int inf = 1e9+1e8;
const int N=1e5+100;
const ll mod=1e9+7;

ll a[N<<2],lazy1[N<<2],lazy2[N<<2],n,m,p;
void push_up(int root){
    a[root]=(a[lrt]+a[rrt])%p;
}
void push_down(int root, int m){
    lazy1[lrt]=(lazy1[root]*lazy1[lrt])%p;
    lazy1[rrt]=(lazy1[root]*lazy1[rrt])%p;
    lazy2[lrt]=((lazy2[lrt]*lazy1[root])%p+lazy2[root])%p;
    lazy2[rrt]=((lazy2[rrt]*lazy1[root])%p+lazy2[root])%p;
    a[lrt]=((a[lrt]*lazy1[root])%p+(lazy2[root]*(m-(m>>1))%p))%p;
    a[rrt]=((a[rrt]*lazy1[root])%p+(lazy2[root]*(m>>1))%p)%p;
    lazy1[root]=1,lazy2[root]=0;
}
void creat(int root, int l, int r){
    if(l==r){
        scanf("%lld",&a[root]);
        a[root]%=p;
        return;
    }
    int mid=l+r>>1;
    creat(lrt, l, mid);
    creat(rrt, mid+1, r);
    push_up(root);
}
void updata1(int root, int l, int r, int L, int R, int c){
    if(l==L && r==R){
        lazy1[root]=(lazy1[root]*(c%p))%p;
        lazy2[root]=(lazy2[root]*(c%p))%p;
        a[root]=(a[root]*(c%p))%p;
        return;
    }
    if(lazy1[root]!=1 || lazy2[root]>0) push_down(root,len);
    int mid=l+r>>1;
    if(R<=mid)updata1(lrt, l, mid, L, R, c);
    else if(L>mid) updata1(rrt, mid+1, r, L, R, c);
    else updata1(lrt, l, mid, L, mid, c) , updata1(rrt, mid+1, r, mid+1, R, c);
    push_up(root);
}
void updata2(int root, int l, int r, int L, int R, int c){
    if(l==L && r==R){
        lazy2[root]=(((lazy2[root]+(c%p))%p))%p;
        a[root]=((a[root]+(c%p)*len)%p)%p;
        return;
    }
    if(lazy1[root]!=1 || lazy2[root]>0) push_down(root,len);
    int mid=l+r>>1;
    if(R<=mid) updata2(lrt, l, mid, L, R, c);
    else if(L>mid) updata2(rrt, mid+1, r, L, R, c);
    else updata2(lrt, l, mid, L, mid, c) , updata2(rrt, mid+1, r, mid+1, R, c);
    push_up(root);
}
ll query(int root, int l, int r, int L, int R){
    if(l==L && r==R){
        return a[root]%p;
    }
    if(lazy1[root]!=1 || lazy2[root]>0) push_down(root,len);
    int mid=l+r>>1;
    if(R<=mid) return query(lrt, l, mid, L, R)%p;
    else if(L>mid) return query(rrt, mid+1, r, L, R)%p;
    else return (query(lrt, l, mid, L, mid)+query(rrt, mid+1, r, mid+1, R))%p;
}
int main(){
    ///ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    cin>>n>>p;
    for(int i=0; i<=4*n; ++i) lazy1[i]=1;
    creat(1,1,n);
    cin>>m;
    int t,c,l,r;
    for(int i=1; i<=m; ++i){
        scanf("%d%d%d",&t,&l,&r);
        if(t==1){
            scanf("%d",&c);
            updata1(1,1,n,l,r,c);
        }
        else if(t==2){
            scanf("%d",&c);
            updata2(1,1,n,l,r,c);
        }
        if(t==3) printf("%lld\n",(query(1,1,n,l,r)+p)%p);
    }
    return 0;
}