【P3373】 【模板】线段树 2 {线段树，模板}

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <algorithm>
  4 #define repo(i,a,b) for(int i = a;i <= b;++i)
  5 #define repi(i,a,b) for(int i = a;i >= b;++i)
  6 using namespace std;
  7 
  8 struct segment_tree{
  9     int l,r;
 10     long long sum,add,mul;//sum为区间和，add为加法懒惰标记，mul为乘法懒惰标记 
 11     #define l(x) tree[x].l
 12     #define r(x) tree[x].r
 13     #define sum(x) tree[x].sum
 14     #define add(x) tree[x].add
 15     #define mul(x) tree[x].mul//节省寿命防止眼瞎的宏定义 
 16 } tree[100002*4];
 17 long long a[100002],n,m,mod,oper,x,y,k;//oper表示操作编号 
 18 
 19 inline int read(){//快读 
 20     long long s=0,w=1;
 21     char ch = getchar();
 22     while(ch < ‘0‘ || ch > ‘9‘){
 23         if(ch == ‘-‘)w = -1;
 24         ch = getchar();
 25     }
 26     while(ch >= ‘0‘ && ch <= ‘9‘){
 27         s = s*10+ch-‘0‘;
 28         ch = getchar();
 29     }
 30     return w*s;
 31 }
 32 
 33 inline void build(int p,int l,int r){//建树 
 34     l(p) = l,r(p) = r;
 35     mul(p) = 1;//别忘了初始化mul(p)，C++11才支持struct内non_static变量的初始化 
 36     if(l==r){sum(p) = a[l] % mod;return;}
 37     int mid = (l+r)/2;
 38     build(p*2,l,mid);
 39     build(p*2+1,mid+1,r);
 40     sum(p) = (sum(p*2) + sum(p*2+1)) % mod;
 41 }
 42 
 43 inline int len(int p){//节省寿命防止眼瞎的函数 
 44     return r(p)-l(p)+1;
 45 }
 46 
 47 inline void spread(int p){
 48     if(add(p) || mul(p)!=1){//如果有标记要传的话 
 49         sum(p*2) = sum(p*2)*mul(p) % mod;
 50         sum(p*2+1) = sum(p*2+1)*mul(p) % mod;
 51         mul(p*2) = mul(p*2)*mul(p) % mod;
 52         mul(p*2+1) = mul(p*2+1)*mul(p) % mod;
 53         add(p*2) = add(p*2)*mul(p) % mod;
 54         add(p*2+1) = add(p*2+1)*mul(p) % mod;//子树的add也会受到乘法的影响,即(a+add)*mul 
 55         mul(p) = 1;//mul(p) down
 56         
 57         sum(p*2) = (sum(p*2)+add(p)*len(p*2)) % mod;
 58         sum(p*2+1) = (sum(p*2+1)+add(p)*len(p*2+1)) % mod;
 59         add(p*2) = (add(p*2)+add(p)) % mod;
 60         add(p*2+1) = (add(p*2+1)+add(p)) % mod;
 61         add(p) = 0;//add(p) down
 62     }
 63 }//除了把p*2和p*2+1各自都写出来,还可以写一个devide_spread()函数
 64  //调用devide_spread(p*2),devide_spread(p*2+1)
 65 
 66 inline void oper_add(int p,int l,int r,int d){//加法操作 
 67     if(l <= l(p) && r >= r(p)){
 68         sum(p) = (sum(p)+(long long)d*len(p)) % mod;
 69         add(p) = (add(p)+d) % mod;
 70         return;
 71     }
 72     spread(p);
 73     int mid = (l(p)+r(p))/2;
 74     if(l <= mid)oper_add(p*2,l,r,d);
 75     if(r > mid)oper_add(p*2+1,l,r,d);
 76     sum(p) = (sum(p*2)+sum(p*2+1)) % mod;
 77 }
 78 
 79 inline void oper_mul(int p,int l,int r,int d){//乘法操作 
 80     if(l <= l(p) && r >= r(p)){
 81         sum(p) = (sum(p)*d) % mod;
 82         add(p) = (add(p)*d) % mod;//同spread()函数里的解释 
 83         mul(p) = (mul(p)*d) % mod;
 84         return;
 85     }
 86     spread(p);
 87     int mid = (l(p)+r(p))/2;
 88     if(l <= mid)oper_mul(p*2,l,r,d);
 89     if(r > mid)oper_mul(p*2+1,l,r,d);
 90     sum(p) = (sum(p*2)+sum(p*2+1)) % mod;
 91 }
 92 
 93 inline long long ask(int p,int l,int r){//询问 
 94     if(l <= l(p) && r >= r(p))return sum(p) % mod;
 95     spread(p);
 96     int mid = (l(p)+r(p))/2;
 97     long long val = 0;
 98     if(l <= mid)val = (val+ask(p*2,l,r)) % mod;
 99     if(r > mid)val = (val+ask(p*2+1,l,r)) % mod;
100     return val;
101 }
102 
103 int main(){
104     n = read(),m = read(),mod = read();
105     repo(i,1,n){
106         a[i] = read();
107     }
108     build(1,1,n);
109     repo(i,1,m){
110         oper = read();
111         if(oper == 1){
112             x = read(),y = read(),k = read();
113             oper_mul(1,x,y,k);
114         }
115         if(oper == 2){
116             x = read(),y = read(),k = read();
117             oper_add(1,x,y,k);
118         }
119         if(oper == 3){
120             x = read(),y = read();
121             printf("%lld\n",ask(1,x,y));
122         }
123     }
124     return 0;
125 }