| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 62228 | Accepted: 19058 | |
| Case Time Limit: 2000MS | ||
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
感觉成段更新好难，lazy数组标记用的很巧妙，以后多做题一定要把它理解透了。
#include <cstdio> #include <iostream> #include <algorithm> #include <cstring> #include <cctype> #include <cmath> #include <cstdlib> #include <vector> #include <queue> #include <set> #include <map> #include <list> #define L long long using namespace std; const int INF=1<<27; const int maxn=500010; L lazy[maxn],sum[maxn]; void push_up(int root) { sum[root]=sum[root*2]+sum[root*2+1]; } void push_down(int root,int l,int r) { if(lazy[root]) { int m=r-l+1; lazy[root*2]+=lazy[root]; lazy[root*2+1]+=lazy[root]; sum[root*2]+=(m-m/2)*lazy[root]; sum[root*2+1]+=(m/2)*lazy[root]; lazy[root]=0; } } void update(int root,int l,int r,int ql,int qr,L v) { if(ql>r||qr<l) return ; if(ql<=l&&qr>=r) { lazy[root]+=v; sum[root]+=(v*(r-l+1)); return ;// } int mid=(l+r)/2; push_down(root,l,r); update(root*2,l,mid,ql,qr,v); update(root*2+1,mid+1,r,ql,qr,v); push_up(root); } L query_sum(int root,int l,int r,int ql,int qr) { if(ql>r||qr<l) return 0; if(ql<=l&&qr>=r) return sum[root]; push_down(root,l,r); int mid=(l+r)/2; return query_sum(root*2,l,mid,ql,qr)+query_sum(root*2+1,mid+1,r,ql,qr); } int main() { int n,Q,i,ql,qr;L v;char op; scanf("%d%d",&n,&Q); //memset(lazy,0,sizeof(lazy)); //memset(sum,0,sizeof(sum)); for(i=1;i<=n;i++) { scanf("%lld",&v); update(1,1,n,i,i,v); } while(Q--) { getchar(); scanf("%c",&op); if(op=='Q') { scanf("%d%d",&ql,&qr); printf("%lld\n",query_sum(1,1,n,ql,qr)); } else { scanf("%d%d%lld",&ql,&qr,&v); update(1,1,n,ql,qr,v); } } return 0; }
POJ 3468-A Simple Problem with Integers（线段树：成段更新，区间求和）,布布扣,bubuko.com
POJ 3468-A Simple Problem with Integers（线段树：成段更新，区间求和）
原文：http://blog.csdn.net/qq_16255321/article/details/38702723