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 }