A Simple Problem with Integers-POJ3468
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 abc" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q ab" 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.
这是线段树模板题,段更新,有一个向下压的过程。
#include<stdio.h> #include<math.h> #include<algorithm> #include<iostream> #include<string.h> #include<stdlib.h> using namespace std; #define N 100100 #define Lson r<<1 #define Rson r<<1|1 struct node { int L,R; long long sum,e; int mid() { return (L+R)/2; } int len() { return R-L+1; } }a[N<<2]; void BuildTree(int r,int L,int R) { a[r].L=L; a[r].R=R; a[r].e=0; if(L==R) { scanf("%lld",&a[r].sum); return; } BuildTree(Lson,L,a[r].mid()); BuildTree(Rson,a[r].mid()+1,R); a[r].sum=a[Lson].sum+a[Rson].sum; } void Down(int r) { a[Lson].sum+=a[Lson].len()*a[r].e; a[Lson].e+=a[r].e; a[Rson].sum+=a[Rson].len()*a[r].e; a[Rson].e+=a[r].e; a[r].e=0; } void Add(int r,int L,int R,int e) { a[r].sum+=(R-L+1)*e; if(a[r].L==L && a[r].R==R) { a[r].e+=e; return; } Down(r); if(R<=a[r].mid()) Add(Lson,L,R,e); else if(L>a[r].mid()) Add(Rson,L,R,e); else { Add(Lson,L,a[r].mid(),e); Add(Rson,a[r].mid()+1,R,e); } } long long Qurry(int r,int L,int R) { if(a[r].L==L && a[r].R==R) { return a[r].sum; } Down(r); if(R<=a[r].mid()) return Qurry(Lson,L,R); else if(L>a[r].mid()) return Qurry(Rson,L,R); else { long long int a1=Qurry(Lson,L,a[r].mid()); long long int a2=Qurry(Rson,a[r].mid()+1,R); return a1+a2; } } int main() { int n,m,d,b,c; while(scanf("%d %d",&n,&m)!=EOF) { BuildTree(1,1,n); char s[10]; while(m--) { scanf("%s",s); if(s[0]=='Q') { scanf("%d %d",&d,&b); printf("%lld\n",Qurry(1,d,b)); } else { scanf("%d %d %d",&d,&b,&c); Add(1,d,b,c); } } } return 0; }