[bzoj 1798][luogu p2023]Seq 线段树Seq

题目大意:

维护一个数列,支持区间乘,区间加,求区间和.

线段树题,对于乘和加操作我们可以维护一个标记.对于乘用乘法分配律分解.

代码如下:

  1 #include <cstdio>
  2 #include <cstdlib>
  3 #include <cstring>
  4 #include <algorithm>
  5 using namespace std;
  6 const int N = 100010;
  7 int inline getint()
  8 {
  9     int num = 0;char ch = getchar();for(;ch < ‘0‘ || ch > ‘9‘;ch = getchar());
 10     for(;ch >= ‘0‘ && ch <= ‘9‘;ch = getchar()) num = (num << 3) + (num << 1) + ch - ‘0‘;
 11     return num;
 12 }
 13 void putint(int x){if(x > 9)putint(x / 10);putchar(x % 10 + ‘0‘);}
 14 int sum[1 + (N << 2)],col1[1 + (N << 2)],col2[1 + (N << 2)],n,m,p;
 15 void pushup(int &x){sum[x] = (sum[x << 1] + sum[x << 1 | 1]) % p;}
 16 void build(int l,int r,int x)
 17 {
 18     col1[x] = 1;
 19     if(l == r){sum[x] = getint() % p;return;}
 20     int mid = l + r >> 1;
 21     if(l <= mid) build(l,mid,x << 1);
 22     if(r > mid) build(mid + 1,r,x << 1 | 1);
 23     pushup(x);
 24 }
 25 void pushdown(int &x,int &l,int &r)
 26 {
 27     if(col1[x] != 1)
 28     {
 29         col1[x << 1] = (long long)col1[x << 1] * col1[x] % p;
 30         col2[x << 1] = (long long)col2[x << 1] * col1[x] % p;
 31         sum[x << 1] = (long long)sum[x << 1] * col1[x] % p;
 32         col1[x << 1 | 1] = (long long)col1[x << 1 | 1] * col1[x] % p;
 33         col2[x << 1 | 1] = (long long)col2[x << 1 | 1] * col1[x] % p;
 34         sum[x << 1 | 1] = (long long)sum[x << 1 | 1] * col1[x] % p;
 35         col1[x] = 1;
 36     }
 37     if(col2[x])
 38     {
 39         col2[x << 1] = (col2[x << 1] + col2[x]) % p;
 40         sum[x << 1] = ((long long)col2[x] * ((l + r >> 1) - l + 1) % p + sum[x << 1]) % p;
 41         col2[x << 1 | 1] = (col2[x << 1 | 1] + col2[x]) % p;
 42         sum[x << 1 | 1] = ((long long)col2[x] * (r - (l + r >> 1)) % p + sum[x << 1 | 1]) % p;
 43         col2[x] = 0;
 44     }
 45 }
 46 void add(int l,int r,int root,int L,int R,int x)
 47 {
 48     if(l >= L && r <= R)
 49     {
 50         col2[root] = (col2[root] + x) % p;
 51         sum[root] = ((long long)(r - l + 1) * x % p + sum[root]) % p;
 52         return;
 53     }
 54     pushdown(root,l,r);
 55     int mid = l + r >> 1;
 56     if(L <= mid) add(l,mid,root << 1,L,R,x);
 57     if(R > mid) add(mid + 1,r,root << 1 | 1,L,R,x);
 58     pushup(root);
 59 }
 60 void mul(int l,int r,int root,int L,int R,int x)
 61 {
 62     if(l >= L && r <= R)
 63     {
 64         col1[root] = (long long)col1[root] * x % p;
 65         col2[root] = (long long)col2[root] * x % p;
 66         sum[root] = (long long)sum[root] * x % p;
 67         return;
 68     }
 69     pushdown(root,l,r);
 70     int mid = l + r >> 1;
 71     if(L <= mid) mul(l,mid,root << 1,L,R,x);
 72     if(R > mid) mul(mid + 1,r,root << 1 | 1,L,R,x);
 73     pushup(root);
 74 }
 75 int query(int l,int r,int root,int L,int R)
 76 {
 77     if(l >= L && r <= R)return sum[root];
 78     pushdown(root,l,r);
 79     int mid = l + r >> 1,ret = 0;
 80     if(L <= mid)ret = query(l,mid,root << 1,L,R);
 81     if(R > mid)ret = (ret + query(mid + 1,r,root << 1 | 1,L,R)) % p;
 82     return ret;
 83 }
 84 int main()
 85 {
 86     n = getint();p = getint();build(1,n,1);
 87     m = getint();
 88     for(int i = 1;i <= m;i++)
 89     {
 90         int opt = getint(),t,g,c;
 91         switch(opt)
 92         {
 93             case 1:
 94                 t = getint();g = getint();c = getint() % p;
 95                 mul(1,n,1,t,g,c);break;
 96             case 2:
 97                 t = getint();g = getint();c = getint() % p;
 98                 add(1,n,1,t,g,c);break;
 99             case 3:
100                 t = getint();g = getint();
101                 putint(query(1,n,1,t,g));putchar(‘\n‘);
102                 break;
103         }
104     }
105     return 0;
106 }