A Simple Problem with Integers
| 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）