POJ_3468 A Simple Problem with Integers 【线段树区间查询+修改】

一、题目

POJ3468

二、分析

裸的线段树区间查询+修改。

三、AC代码

  1 #include <cstdio>
  2 #include <iostream>
  3 #include <algorithm>
  4 
  5 using namespace std;
  6 
  7 #define ll long long
  8 #define lson l, mid, rt << 1
  9 #define rson mid + 1, r, rt << 1 | 1
 10 const int MAXN = 1e5 + 3;
 11 ll add[MAXN << 2], sum[MAXN << 2];
 12 
 13 void PushUp(int rt)
 14 {
 15     sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
 16     return;
 17 }
 18 
 19 void PushDown(int ln, int rn, int rt)
 20 {
 21     if(add[rt])
 22     {
 23         sum[rt << 1] += (ll) ln * add[rt];
 24         sum[rt << 1 | 1] += (ll) rn * add[rt];
 25         add[rt << 1] += add[rt];
 26         add[rt << 1 | 1] += add[rt];
 27         add[rt] = 0;
 28     }
 29     return;
 30 }
 31 
 32 void Build(int l, int r, int rt)
 33 {
 34     add[rt] = 0;
 35     if(l == r)
 36     {
 37         scanf("%lld", &sum[rt]);
 38         return;
 39     }
 40     int mid = (l + r) >> 1;
 41     Build(lson);
 42     Build(rson);
 43     PushUp(rt);
 44     return;
 45 }
 46 
 47 void Update(int L, int R, int c, int l, int r, int rt)
 48 {
 49     if(L <= l && r <= R)
 50     {
 51         add[rt] += c;
 52         sum[rt] += (ll)(r - l + 1) * c;
 53         return;
 54     }
 55     int mid = (l + r) >> 1;
 56     PushDown(mid - l + 1, r - mid, rt);
 57     if(L <= mid)
 58     {
 59         Update(L, R, c, lson);
 60     }
 61     if(R > mid)
 62     {
 63         Update(L, R, c, rson);
 64     }
 65     PushUp(rt);
 66     return;
 67 }
 68 
 69 ll Query(int L, int R, int l, int r, int rt)
 70 {
 71     if(L <= l && r <= R)
 72         return sum[rt];
 73     int mid = (l + r) >> 1;
 74     PushDown(mid - l + 1, r - mid, rt);
 75     ll ans = 0;
 76     if(L <= mid)
 77     {
 78         ans += Query(L, R, lson);
 79     }
 80     if(R > mid)
 81     {
 82         ans += Query(L, R, rson);
 83     }
 84     return ans;
 85 }
 86 
 87 int main()
 88 {
 89     //freopen("input.txt", "r", stdin);
 90     int N, Q;
 91     scanf("%d %d", &N, &Q);
 92     Build(1, N, 1);
 93     char s[2];
 94     for(int i = 0; i < Q; i++)
 95     {
 96         int L, R, C;
 97         scanf("%s", s);
 98         if(s[0] == 'Q')
 99         {
100             scanf("%d %d", &L, &R);
101             printf("%I64d\n", Query(L, R, 1, N, 1));
102         }
103         else
104         {
105             scanf("%d %d %d", &L, &R, &C);
106             Update(L, R, C, 1, N, 1);
107         }
108     }
109 }
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posted @ 2019-03-07 16:54  Dybala21  阅读(93)  评论(0)  编辑  收藏  举报