POJ 3468 A Simple Problem with Integers

Posted 2020-09-27 EricJeffrey

线段树区间更新模板题，加个延迟标记即可。

注意:区间大小是r-l+1，query，更新时要更新子节点，同时注意+=与=！

discuss里面有数据，wa的可以看看。。。

代码：

  1 #include <map>
  2 #include <set>
  3 #include <cmath>
  4 #include <queue>
  5 #include <stack>
  6 #include <cstdio>
  7 #include <string>
  8 #include <vector>
  9 #include <cstdlib>
 10 #include <cstring>
 11 #include <sstream>
 12 #include <iostream>
 13 #include <algorithm>
 14 #include <functional>
 15 using namespace std;
 16 #define rep(i,a,n) for (int i=a;i<n;i++)
 17 #define per(i,a,n) for (int i=n-1;i>=a;i--)
 18 #define all(x) (x).begin(),(x).end()
 19 #define pb push_back
 20 #define mp make_pair
 21 #define lson l,m,rt<<1
 22 #define rson m+1,r,rt<<1|1
 23 typedef long long ll;
 24 typedef vector<int> VI;
 25 typedef pair<int, int> PII;
 26 const ll MOD = 1e9 + 7;
 27 const int inf = 0x3f3f3f3f;
 28 //head
 29 #define maxn 500010
 30 struct node{
 31     int left, right;
 32     long long val, lazy;
 33     node() {}
 34     node(int l, int r, long long v, long long la): left(l), right(r), val(v), lazy(la){}
 35 };
 36 node tree[maxn];
 37 int a[maxn], n;
 38 
 39 void build(int l, int r, int root){
 40     tree[root] = node(l, r, 0, 0);
 41     if(l == r)
 42         tree[root].val = a[l];
 43     else{
 44         int mid = (l + r) / 2;
 45         build(l, mid, root << 1);
 46         build(mid + 1, r, root << 1 | 1);
 47         tree[root].val = tree[root << 1].val + tree[root << 1 | 1].val;
 48     }
 49 }
 50 void update(int ul, int ur, int v, int l, int r, int k){
 51     if(ul == l && ur == r){
 52         tree[k].lazy += v;
 53         return;
 54     }
 55     if(ul > r || ur < l) 
 56         return;
 57     
 58     tree[k].val += tree[k].lazy * (tree[k].right - tree[k].left + 1);
 59     tree[k << 1].lazy += tree[k].lazy;
 60     tree[k << 1 | 1].lazy += tree[k].lazy;
 61     tree[k].lazy = 0;
 62     
 63     tree[k].val += v * (ur - ul + 1);
 64     int mid = (l + r) / 2;
 65     if(ur <= mid){ 
 66         update(ul, ur, v, l, mid, k << 1);
 67     }
 68     else if(ul <= mid){
 69         update(ul, mid, v, l, mid, k << 1);
 70         update(mid + 1, ur, v, mid + 1, r, k << 1 | 1);
 71     } 
 72     else{
 73         update(ul, ur, v, mid + 1, r, k << 1 | 1);
 74     }
 75 }
 76 long long query(int ql, int qr, int l, int r, int k){
 77     if(ql == l && qr == r){
 78         tree[k].val += tree[k].lazy * (tree[k].right - tree[k].left + 1);
 79         tree[k << 1].lazy += tree[k].lazy;
 80         tree[k << 1 | 1].lazy += tree[k].lazy;
 81         tree[k].lazy = 0;
 82         return tree[k].val;
 83     } 
 84         
 85 
 86     tree[k].val += tree[k].lazy * (tree[k].right - tree[k].left + 1);
 87     tree[k << 1].lazy += tree[k].lazy;
 88     tree[k << 1 | 1].lazy += tree[k].lazy;
 89     tree[k].lazy = 0;
 90     
 91     int mid = (l + r) / 2;
 92     if(qr <= mid){
 93         return query(ql, qr, l, mid, k << 1);
 94     }
 95     else if(ql <= mid){
 96         return query(ql, mid, l, mid, k << 1) + query(mid + 1, qr, mid + 1, r, k << 1 | 1);
 97     }
 98     else 
 99         return query(ql, qr, mid + 1, r, k << 1 | 1);
100 }
101 
102 int main(){
103     memset(a, 0, sizeof(a));
104     memset(tree, 0 ,sizeof(tree));
105     int q;
106     scanf("%d %d", &n, &q);
107     for(int i = 0; i < n; i++)
108         scanf("%lld", &a[i]); 
109     build(0, n - 1, 1);
110     while(q--){
111         char ch;
112         int tma, tmb;
113         scanf(" %c %d %d", &ch, &tma, &tmb);
114         tma--; tmb--;
115         if(ch == ‘Q‘){
116             //debug
117             //for(int i = 0; i < 28; i++) printf("%d %d %d\n", i, tree[i].val, tree[i].lazy);
118             printf("%lld\n", query(tma, tmb, 0, n - 1, 1));    
119         }
120         else if(ch == ‘C‘){
121             int tmc;
122             scanf("%d", &tmc);
123             update(tma, tmb, tmc, 0, n - 1, 1);
124             //debug
125             //for(int i = 0; i < 28; i++) printf("%d %d %d\n", i, tree[i].val, tree[i].lazy);
126         }
127         //debug
128         //if(q == 3){
129         //for(int i = 0; i < 28; i++) printf("%d %d %d\n", i, tree[i].val, tree[i].lazy);
130         //} 
131     }
132 }

题目：

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 110849		Accepted: 34520
Case Time Limit: 2000MS

Description

You have N integers, A₁, A₂, ... , A_N. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.

Input

The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A₁, A₂, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a+1, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a, A_a+1, ... , A_b.

Output

You need to answer all Q commands in order. One answer in a line.

Sample Input

10 5
1 2 3 4 5 6 7 8 9 10
Q 4 4
Q 1 10
Q 2 4
C 3 6 3
Q 2 4

Sample Output

4
55
9
15

Hint

The sums may exceed the range of 32-bit integers.