線段樹-----C - A Simple Problem with Integers
You have N integers, A1, A2, … , AN. 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, … , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
“C a b c” means adding c to each of Aa, Aa+1, … , Ab. -10000 ≤ c ≤ 10000.
“Q a b” means querying the sum of Aa, Aa+1, … , Ab.
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.
線段樹模板題,區間更新時候那個延遲標記要注意,求區間和,每次val值要加上區間長度乘以addval值,求區間最大值,每次val值直接加上addval值即可
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<map> #include<queue> using namespace std; typedef long long LL; const int inf = 0x3f3f3f3f; #define pb push_back #define mp make_pair #define fi first #define se second const int N = 2 * 1e5 + 5; typedef struct Node{ LL val,addval; }Node; Node tree[N << 2]; LL a[N]; void build(int root,int l,int r) { tree[root].addval = 0; if(l == r){ tree[root].val = a[l]; return ; } int mid = (l + r) / 2; build(root * 2 + 1,l,mid); build(root * 2 + 2,mid + 1,r); tree[root].val = tree[root * 2 + 1].val + tree[root * 2 + 2].val; } void pushdown(int root,int m) { if(tree[root].addval != 0){ tree[root * 2 + 1].addval += tree[root].addval; tree[root * 2 + 2].addval += tree[root].addval; tree[root * 2 + 1].val += (m - (m >> 1)) * tree[root].addval; tree[root * 2 + 2].val += (m >> 1) * tree[root].addval; tree[root].addval = 0; } } void update(int root,int l,int r,int pl,int pr,int val) { if(r < pl || l > pr){ return ; } if(l >= pl && r <= pr){ tree[root].addval += val; tree[root].val += (r - l + 1) * val; return ; } pushdown(root,r - l + 1); int mid = (l + r) / 2; if(pl <= mid){ update(root * 2 + 1,l,mid,pl,pr,val); } if(pr > mid){ update(root * 2 + 2,mid + 1,r,pl,pr,val); } tree[root].val = tree[root * 2 + 1].val + tree[root * 2 + 2].val; } LL query(int root,int l,int r,int pl,int pr) { if(r < pl || l > pr){ return 0; } if(l >= pl && r <= pr){ return tree[root].val; } pushdown(root,r - l + 1); int mid = (l + r) / 2; LL sum = 0; if(pl <= mid){ sum += query(root * 2 + 1,l,mid,pl,pr); } if(pr > mid){ sum += query(root * 2 + 2,mid + 1,r,pl,pr); } return sum; } int main() { int n,q; while(~scanf("%d %d",&n,&q)) { for(int i = 1;i <= n;++i){ scanf("%lld",&a[i]); } build(1,1,n); for(int i = 0;i < q;++i){ char s[5]; scanf("%s",s); if(s[0] == 'Q'){ int x,y; scanf("%d %d",&x,&y); printf("%lld\n",query(1,1,n,x,y)); }else{ int x,y,z; scanf("%d %d %d",&x,&y,&z); update(1,1,n,x,y,z); } } } return 0; }