标签：i++ blank -- exce iostream put line number inpu
Description
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
Source
给定一个数列有n个数以及q个操作
有两种操作：
c：给出a,b,c 对a，b区间的数同时加上c
q：给出a,b 求a,b区间的和
如果对每个节点，维护对应区间的和，可以在O（logn）时间内求出任意区间的和，但是不能够高效的实现对一个区间同时加一个值，因为需要对其所有关联节点进行更新
为了保持线段树的高效，对于每个节点维护两个数据：
a 给这个节点对应的区间内同时加上一个值
b在这个节点对应的区间内除去a之外其他值的和
#include <stdio.h> #include <algorithm> #include <cmath> #include <cstring> #include <deque> #include <iomanip> #include <iostream> #include <list> #include <map> #include <queue> #include <set> #include <utility> #include <vector> #define mem(arr, num) memset(arr, 0, sizeof(arr)) #define _for(i, a, b) for (int i = a; i <= b; i++) #define __for(i, a, b) for (int i = a; i >= b; i--) #define IO \ ios::sync_with_stdio(false); cin.tie(0); cout.tie(0); using namespace std; typedef long long ll; const ll inf = 0x3f3f3f3f; const double EPS = 1e-10; const ll mod = 1000000007LL; const int N = 1 << 19; ll data[N],datb[N]; int n,q; void build(int l, int r, int k) { data[k] = datb[k] = 0; if(l==r) { scanf("%d",&datb[k]);return ; } build(l,(l+r)/2,k<<1); build((l+r)/2+1,r,k<<1|1); datb[k] = datb[k<<1] + datb[k<<1|1]; } void update(int a,int b,int k,int l,int r,int dat) { if(a <= l && b >= r) { data[k] += dat; } else if(a <= r && b >= l) { datb[k] += (min(b,r) - max(a,l) + 1) * x; update(a,b,k<<1,l,(l+r)/2,dat); update(a,b,k<<1|1,(l+r)/2+1,r,dat); } } ll query(int a,int b,int k,int l,int r) { if(b < l || a > r) return 0; else if(a <= l && b >= r) return data[k] * (r-l+1) + datb[k]; else { ll res = (min(b,r) - max(a,l) + 1) *data[k]; res += query(a,b,k<<1,l,(l+r)/2); res += query(a,b,k<<1|1,(l+r)/2+1); return res; } } int main() { scanf("%d%d",&n,&q); build(1,n,1); char op; int a,b,c; _for(i, 1, q) { getchar(); scanf("%c",&op); if(op == ‘Q‘) { scanf("%d%d",&a,&b); printf("%lld\n",query(a,b,1,1,n)); } else { scanf("%d%d%d",&a,&b,&c); update(a,b,1,1,n,c); } } return 0; }
POJ 3468 A Simple Problem with Integers（线段树区间求和）
标签：i++ blank -- exce iostream put line number inpu
原文地址：https://www.cnblogs.com/GHzz/p/8920155.html
