线段树(成段更新) POJ 3468 A Simple Problem with Integers

题目传送门

  1 /*
  2     线段树-成段更新：裸题，成段增减，区间求和
  3     注意：开long long:)
  4 */
  5 #include <cstdio>
  6 #include <iostream>
  7 #include <algorithm>
  8 #include <cstring>
  9 #include <cmath>
 10 using namespace std;
 11
 12 #define lson l, mid, rt << 1
 13 #define rson mid + 1, r, rt << 1 | 1
 14 #define LL long long
 15
 16 const int MAXN = 1e5 + 10;
 17 const int INF = 0x3f3f3f3f;
 18 struct Node
 19 {
 20     LL sum, add;
 21 }node[MAXN << 2];
 22
 23 void push_up(int rt)
 24 {
 25     node[rt].sum = node[rt<<1].sum + node[rt<<1|1].sum;
 26 }
 27
 28 void push_down(int rt, int c)
 29 {
 30     if (node[rt].add)
 31     {
 32         node[rt<<1].add += node[rt].add;
 33         node[rt<<1|1].add += node[rt].add;
 34         node[rt<<1].sum += node[rt].add * (c - (c >> 1));
 35         node[rt<<1|1].sum += node[rt].add * (c >> 1);
 36         node[rt].add = 0;
 37     }
 38 }
 39
 40 void build(int l, int r, int rt)
 41 {
 42     node[rt].add = 0;
 43     if (l == r)    {scanf ("%I64d", &node[rt].sum);    return ;}
 44     int mid = (l + r) >> 1;
 45     build (lson);    build (rson);
 46
 47     push_up (rt);
 48 }
 49
 50 void updata(int ql, int qr, int c, int l, int r, int rt)
 51 {
 52     if (ql <= l && r <= qr)    {node[rt].sum += (LL) c * (r - l + 1);    node[rt].add += c;    return ;}
 53
 54     push_down (rt, r - l + 1);
 55
 56     int mid = (l + r) >> 1;
 57     if (ql <= mid)    updata (ql, qr, c, lson);
 58     if (qr > mid)    updata (ql, qr, c, rson);
 59
 60     push_up (rt);
 61 }
 62
 63 LL query(int ql, int qr, int l, int r, int rt)
 64 {
 65     if (ql <= l && r <= qr)    return node[rt].sum;
 66
 67     push_down (rt, r - l + 1);
 68
 69     int mid = (l + r) >> 1;        LL ans = 0;
 70     if (ql <= mid)    ans += query (ql, qr, lson);
 71     if (qr > mid)    ans += query (ql, qr, rson);
 72
 73     return ans;
 74 }
 75
 76 int main(void)        //POJ 3468 A Simple Problem with Integers
 77 {
 78     //freopen ("POJ_3468.in", "r", stdin);
 79
 80     int n, q;
 81     while (scanf ("%d%d", &n, &q) == 2)
 82     {
 83         build (1, n, 1);
 84         while (q--)
 85         {
 86             char s[3];    int ql, qr, c;
 87             scanf ("%s", &s);
 88             if (s[0] == ‘Q‘)
 89             {
 90                 scanf ("%d%d", &ql, &qr);
 91                 printf ("%I64d\n", query (ql, qr, 1, n, 1));
 92             }
 93             else
 94             {
 95                 scanf ("%d%d%d", &ql, &qr, &c);
 96                 updata (ql, qr, c, 1, n, 1);
 97             }
 98         }
 99     }
100
101     return 0;
102 }