https://www.luogu.com.cn/problem/P3373
题意:给出n个数,m次操作,模q , ai , 三种操作:1、l , r , k 区间乘k , 2 、 l , r , k , 区间加k , 3 、l , r 询问区间和。
解法:线段树两个标记乘法标记mul和加法标记plu。pushdown下放加法有点特殊,需要自己身上的加法标记乘以父亲的乘法标记再加上
父亲的加法标记。
#include<bits/stdc++.h> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <iostream> #include <string> #include <stdio.h> #include <queue> #include <stack> #include <map> #include <set> #include <string.h> #include <vector> typedef long long ll ; #define int ll #define mod 1000000007 #define gcd __gcd #define rep(i , j , n) for(int i = j ; i <= n ; i++) #define red(i , n , j) for(int i = n ; i >= j ; i--) #define ME(x , y) memset(x , y , sizeof(x)) //ll lcm(ll a , ll b){return a*b/gcd(a,b);} ll quickpow(ll a , ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;b>>=1,a=a*a%mod;}return ans;} //int euler1(int x){int ans=x;for(int i=2;i*i<=x;i++)if(x%i==0){ans-=ans/i;while(x%i==0)x/=i;}if(x>1)ans-=ans/x;return ans;} //const int N = 1e7+9; int vis[n],prime[n],phi[N];int euler2(int n){ME(vis,true);int len=1;rep(i,2,n){if(vis[i]){prime[len++]=i,phi[i]=i-1;}for(int j=1;j<len&&prime[j]*i<=n;j++){vis[i*prime[j]]=0;if(i%prime[j]==0){phi[i*prime[j]]=phi[i]*prime[j];break;}else{phi[i*prime[j]]=phi[i]*phi[prime[j]];}}}return len} #define SC scanf #define INF 0x3f3f3f3f #define PI acos(-1) #define pii pair<int,int> #define fi first #define se second #define lson l,mid,root<<1 #define rson mid+1,r,root<<1|1 using namespace std; const int N = 1e6+100; const int maxn = 1e5+9; int n , m , p ; int a[maxn]; struct node{ int l , r , sum , mul , plu; }tree[maxn<<2]; void pushup(int root){ tree[root].sum = (tree[root<<1].sum + tree[root<<1|1].sum) % p; } void pushdown(int root){ int k1 = tree[root].mul , k2 = tree[root].plu; tree[root<<1].sum =(k1 * tree[root<<1].sum + k2*(tree[root<<1].r - tree[root<<1].l + 1))% p; tree[root<<1|1].sum = (k1 * tree[root<<1|1].sum + k2*(tree[root<<1|1].r - tree[root<<1|1].l + 1))% p; tree[root<<1].mul = tree[root<<1].mul * k1 % p ; tree[root<<1|1].mul = tree[root<<1|1].mul * k1 % p ; tree[root<<1].plu = (tree[root<<1].plu*k1 + k2) % p ; tree[root<<1|1].plu = (tree[root<<1|1].plu*k1 + k2) % p ; tree[root].mul =1 ; tree[root].plu = 0; } void build(int l , int r , int root){ tree[root].l = l , tree[root].r = r , tree[root].mul = 1 , tree[root].plu = 0; if(l == r){ tree[root].sum = a[l] % p ; return ; } int mid = (l + r) >> 1; build(lson); build(rson); pushup(root); } void update_mul(int l , int r , int root , int val){ if(tree[root].l >= l && tree[root].r <= r){ tree[root].sum = tree[root].sum * val % p; tree[root].mul = tree[root].mul * val % p; tree[root].plu = tree[root].plu * val % p; return ; } pushdown(root); int mid = (tree[root].l + tree[root].r) >> 1; if(mid >= l) update_mul(l , r , root<<1 , val); if(mid < r) update_mul(l , r , root<<1|1 , val); pushup(root); } void update_plu(int l , int r , int root , int val){ if(tree[root].l >= l && tree[root].r <= r){ tree[root].sum = (tree[root].sum + (tree[root].r - tree[root].l + 1) * val) % p ; tree[root].plu = (tree[root].plu + val) % p ; return ; } pushdown(root); int mid = (tree[root].l + tree[root].r) >> 1 ; if(mid >= l) update_plu(l , r , root<<1 , val); if(mid < r) update_plu(l , r , root<<1|1 , val); pushup(root); } int query(int l , int r , int root){ int sum = 0 ; if(tree[root].l >= l && tree[root].r <= r){ return tree[root].sum % p ; } pushdown(root); int mid = (tree[root].l + tree[root].r) >> 1 ; if(mid >= l) sum = (sum + query(l , r , root<<1)) % p; if(mid < r) sum = (sum + query(l , r , root<<1|1)) % p; return sum % p ; } void solve(){ scanf("%lld%lld%lld" , &n , &m , &p); rep(i , 1 , n) scanf("%lld" , &a[i]); build(1 , n , 1); rep(i , 1 , m){ int t , l , r , val; scanf("%lld%lld%lld" , &t , &l , &r); if(t == 1){ scanf("%lld" , &val); val %= p; update_mul(l , r , 1 , val); }else if(t == 2){ scanf("%lld" , &val); val %= p ; update_plu(l , r , 1 , val); } else{ cout << query(l , r , 1) << endl; } } } signed main() { //ios::sync_with_stdio(false); //cin.tie(0); cout.tie(0); //int t ; //cin >> t ; //while(t--){ solve(); //} }