Yuanfang is puzzled with the question below:
There are n integers, a1, a2, …, an. The initial values of them are 0. There are four kinds of operations.
Operation 1: Add c to each number between ax and ay inclusive. In other words, do transformation ak<---ak+c, k = x,x+1,…,y.
Operation 2: Multiply c to each number between ax and ay inclusive. In other words, do transformation ak<---ak×c, k = x,x+1,…,y.
Operation 3: Change the numbers between ax and ay to c, inclusive. In other words, do transformation ak<---c, k = x,x+1,…,y.
Operation 4: Get the sum of p power among the numbers between ax and ay inclusive. In other words, get the result of axp+ax+1p+…+ay p.
Yuanfang has no idea of how to do it. So he wants to ask you to help him.
区间加、乘、改、求和，线段树，推出式子即可。
1 #include<stdio.h>
2 #include<string.h>
3 const int mod=10007;
4 const int maxm=100005;
5
6 int st[4][maxm<<2],add[maxm<<2],mul[maxm<<2],cha[maxm<<2];
7
8 void build(int o,int l,int r){
9 st[1][o]=st[2][o]=st[3][o]=0;
10 add[o]=cha[o]=0;
11 mul[o]=1;
12 if(l==r)return;
13 int m=l+((r-l)>>1);
14 build(o<<1,l,m);
15 build(o<<1|1,m+1,r);
16 }
17
18 void pushdown(int o,int l,int r){
19 if(cha[o]){
20 cha[o<<1]=cha[o<<1|1]=cha[o];
21 add[o<<1]=add[o<<1|1]=0;
22 mul[o<<1]=mul[o<<1|1]=1;
23 int m=l+((r-l)>>1);
24 st[1][o<<1]=cha[o]*(m-l+1)%mod;
25 st[1][o<<1|1]=cha[o]*(r-m)%mod;
26 st[2][o<<1]=(cha[o]*cha[o]%mod)*(m-l+1)%mod;
27 st[2][o<<1|1]=(cha[o]*cha[o]%mod)*(r-m)%mod;
28 st[3][o<<1]=((cha[o]*cha[o]%mod)*cha[o]%mod)*(m-l+1)%mod;
29 st[3][o<<1|1]=((cha[o]*cha[o]%mod)*cha[o]%mod)*(r-m)%mod;
30 cha[o]=0;
31 }
32 if(add[o]||(mul[o]!=1)){
33 int m=l+((r-l)>>1);
34 add[o<<1]=(add[o<<1]*mul[o]%mod+add[o])%mod;
35 mul[o<<1]=mul[o<<1]*mul[o]%mod;
36 st[3][o<<1]=(
37 ((st[3][o<<1]*mul[o]%mod)*mul[o]%mod)*mul[o]%mod
38 +(((st[2][o<<1]*mul[o]%mod)*mul[o]%mod)*add[o]%mod)*3%mod
39 +(((st[1][o<<1]*mul[o]%mod)*add[o]%mod)*add[o]%mod)*3%mod
40 +((add[o]*add[o]%mod)*add[o]%mod)*(m-l+1)%mod
41 )%mod;
42 st[2][o<<1]=(
43 (st[2][o<<1]*mul[o]%mod)*mul[o]%mod
44 +((st[1][o<<1]*mul[o]%mod)*add[o]%mod)*2%mod
45 +(add[o]*add[o]%mod)*(m-l+1)%mod
46 )%mod;
47 st[1][o<<1]=(
48 st[1][o<<1]*mul[o]%mod+add[o]*(m-l+1)%mod
49 )%mod;
50 add[o<<1|1]=(add[o<<1|1]*mul[o]%mod+add[o])%mod;
51 mul[o<<1|1]=mul[o<<1|1]*mul[o]%mod;
52 st[3][o<<1|1]=(
53 ((st[3][o<<1|1]*mul[o]%mod)*mul[o]%mod)*mul[o]%mod
54 +(((st[2][o<<1|1]*mul[o]%mod)*mul[o]%mod)*add[o]%mod)*3%mod
55 +(((st[1][o<<1|1]*mul[o]%mod)*add[o]%mod)*add[o]%mod)*3%mod
56 +((add[o]*add[o]%mod)*add[o]%mod)*(r-m)%mod
57 )%mod;
58 st[2][o<<1|1]=(
59 (st[2][o<<1|1]*mul[o]%mod)*mul[o]%mod
60 +((st[1][o<<1|1]*mul[o]%mod)*add[o]%mod)*2%mod
61 +(add[o]*add[o]%mod)*(r-m)%mod
62 )%mod;
63 st[1][o<<1|1]=(
64 st[1][o<<1|1]*mul[o]%mod+add[o]*(r-m)%mod
65 )%mod;
66 add[o]=0;
67 mul[o]=1;
68 }
69 }
70
71 void update3(int o,int l,int r,int ql,int qr,int c){
72 if(ql<=l&&qr>=r){
73 st[1][o]=c*(r-l+1)%mod;
74 st[2][o]=(c*c%mod)*(r-l+1)%mod;
75 st[3][o]=((c*c%mod)*c%mod)*(r-l+1)%mod;
76 cha[o]=c;
77 add[o]=0;
78 mul[o]=1;
79 return;
80 }
81 pushdown(o,l,r);
82 int m=l+((r-l)>>1);
83 if(ql<=m)update3(o<<1,l,m,ql,qr,c);
84 if(qr>=m+1)update3(o<<1|1,m+1,r,ql,qr,c);
85 st[1][o]=(st[1][o<<1]+st[1][o<<1|1])%mod;
86 st[2][o]=(st[2][o<<1]+st[2][o<<1|1])%mod;
87 st[3][o]=(st[3][o<<1]+st[3][o<<1|1])%mod;
88 }
89
90 void update1(int o,int l,int r,int ql,int qr,int c){
91 if(ql<=l&&qr>=r){
92 add[o]=(add[o]+c)%mod;
93 st[3][o]=(st[3][o]+(c*st[2][o]%mod)*3%mod+((st[1][o]*c%mod)*c%mod)*3%mod+((c*c%mod)*c%mod)*(r-l+1)%mod)%mod;
94 st[2][o]=(st[2][o]+(c*st[1][o]%mod)*2%mod+(c*c%mod)*(r-l+1)%mod)%mod;
95 st[1][o]=(st[1][o]+c*(r-l+1)%mod)%mod;
96 return;
97 }
98 pushdown(o,l,r);
99 int m=l+((r-l)>>1);
100 if(ql<=m)update1(o<<1,l,m,ql,qr,c);
101 if(qr>=m+1)update1(o<<1|1,m+1,r,ql,qr,c);
102 st[1][o]=(st[1][o<<1]+st[1][o<<1|1])%mod;
103 st[2][o]=(st[2][o<<1]+st[2][o<<1|1])%mod;
104 st[3][o]=(st[3][o<<1]+st[3][o<<1|1])%mod;
105 }
106
107 void update2(int o,int l,int r,int ql,int qr,int c){
108 if(ql<=l&&qr>=r){
109 add[o]=add[o]*c%mod;
110 mul[o]=mul[o]*c%mod;
111 st[3][o]=((st[3][o]*c%mod)*c%mod)*c%mod;
112 st[2][o]=(st[2][o]*c%mod)*c%mod;
113 st[1][o]=st[1][o]*c%mod;
114 return;
115 }
116 pushdown(o,l,r);
117 int m=l+((r-l)>>1);
118 if(ql<=m)update2(o<<1,l,m,ql,qr,c);
119 if(qr>=m+1)update2(o<<1|1,m+1,r,ql,qr,c);
120 st[1][o]=(st[1][o<<1]+st[1][o<<1|1])%mod;
121 st[2][o]=(st[2][o<<1]+st[2][o<<1|1])%mod;
122 st[3][o]=(st[3][o<<1]+st[3][o<<1|1])%mod;
123 }
124
125 int query(int o,int l,int r,int ql,int qr,int p){
126 if(ql<=l&&qr>=r){
127 return st[p][o];
128 }
129 pushdown(o,l,r);
130 int m=l+((r-l)>>1);
131 int ans=0;
132 if(ql<=m)ans=(ans+query(o<<1,l,m,ql,qr,p))%mod;
133 if(qr>=m+1)ans=(ans+query(o<<1|1,m+1,r,ql,qr,p))%mod;
134 return ans;
135 }
136
137 int main(){
138 int n,m;
139 while(scanf("%d%d",&n,&m)!=EOF&&n+m){
140 build(1,1,n);
141 for(int i=1;i<=m;++i){
142 int t,l,r,p;
143 scanf("%d%d%d%d",&t,&l,&r,&p);
144 if(t==1)update1(1,1,n,l,r,p);
145 else if(t==2)update2(1,1,n,l,r,p);
146 else if(t==3)update3(1,1,n,l,r,p);
147 else if(t==4)printf("%d\n",query(1,1,n,l,r,p));
148 }
149 }
150 return 0;
151 }
来源：https://www.cnblogs.com/cenariusxz/p/6598413.html