题目链接：https://www.luogu.com.cn/problem/P3373
本题思路主要也是借鉴了洛谷题解区的方法
通过建立两个标签，tag1用于加法记忆，tag2用于乘法记忆
关键是注意两个运算标签的处理 当个简单模板康康
1 #include<bits/stdc++.h>
2 #define ll long long 3 using namespace std; 4 const int maxn=1e5+10;
5 int n,w,m; 6 int temp[4*maxn];
7 struct Segment_Tree 8 { 9 ll val; 10 ll tag1; 11 ll tag2; 12 }data[maxn*4];
13 inline void push_up(int root) 14 { 15 data[root].val=(data[root<<1].val+data[root<<1|1].val)%m; 16 }
17 void build(int left,int right,int root)//root为线段树下标,left为数组下标 18 { 19 data[root].tag1=0; 20 data[root].tag2=1; 21 if(left == right){ 22 data[root].val=temp[left]; 23 return ; 24 } 25 int mid=(left+right)>>1; 26 build(left,mid,root<<1); 27 build(mid+1,right,root<<1|1); 28 data[root].val=data[root<<1].val+data[root<<1|1].val; 29 data[root].val%=m; 30 } 31 32 inline void push_down(int left,int right,int root) 33 { 34 int mid=(left+right)>>1; 35 data[root<<1].val=(data[root<<1].val*data[root].tag2+(mid-left+1)*data[root].tag1)%m; 36 data[root<<1|1].val=(data[root<<1|1].val*data[root].tag2+(right-mid)*data[root].tag1)%m; 37 data[root<<1].tag2=(data[root].tag2*data[root<<1].tag2)%m; 38 data[root<<1|1].tag2=(data[root].tag2*data[root<<1|1].tag2)%m; 39 data[root<<1].tag1=(data[root<<1].tag1*data[root].tag2+data[root].tag1)%m; 40 data[root<<1|1].tag1=(data[root<<1|1].tag1*data[root].tag2+data[root].tag1)%m; 41 data[root].tag1=0; 42 data[root].tag2=1; 43 }
44 void add(int left,int right,int root,int x,int y,int v) 45 { 46 if(x > right || y < left) return ; 47 if(x <= left && y >= right){//注意符号 48 data[root].val=(data[root].val+(right-left+1)*v)%m; 49 data[root].tag1=(data[root].tag1+v)%m; 50 return ; 51 }
52 push_down(left,right,root); 53 int mid=(left+right)>>1; 54 add(left,mid,root<<1,x,y,v); 55 add(mid+1,right,root<<1|1,x,y,v); 56 push_up(root);// 57 }
58 void multiply(int left,int right,int root,int x,int y,int v) 59 { 60 if(x > right || y < left) return ; 61 if(x <= left && y >= right){ 62 data[root].val=(data[root].val*v)%m; 63 data[root].tag2=(data[root].tag2*v)%m; 64 data[root].tag1=(data[root].tag1*v)%m;//加的也要乘 65 return ; 66 } 67 push_down(left,right,root); 68 int mid=(left+right)>>1; 69 multiply(left,mid,root<<1,x,y,v); 70 multiply(mid+1,right,root<<1|1,x,y,v); 71 push_up(root); 72 }
73 ll query(int left,int right,int root,int x,int y) 74 { 75 if(x > right || y < left) return 0; 76 if(x <= left && y >= right){ 77 return data[root].val; 78 } 79 int mid=(left+right)>>1; 80 push_down(left,right,root); 81 return (query(left,mid,root<<1,x,y)+query(mid+1,right,root<<1|1,x,y))%m; 82 }
83 int main(){ 84 cin >> n >> w >> m; 85 for(int i = 1 ; i <= n ; i++){ 86 cin >> temp[i]; 87 } 88 build(1,n,1); 89 int flag,a,b,v; 90 for(int i = 1 ; i <= w ; i++){ 91 scanf("%d",&flag); 92 if(flag == 1){ 93 scanf("%d%d%d",&a,&b,&v); 94 multiply(1,n,1,a,b,v); 95 }else if(flag == 2){ 96 scanf("%d%d%d",&a,&b,&v); 97 add(1,n,1,a,b,v); 98 }else{ 99 scanf("%d%d",&a,&b); 100 printf("%lld ",query(1,n,1,a,b)%m); 101 } 102 } 103 return 0; 104 } 105 106