线段树成段更新——POJ 3468
对应POJ题目：点击打开链接
A Simple Problem with IntegersCrawling in process...Crawling failedTime Limit:5000MS Memory Limit:131072KB 64bit IO Format:%I64d & %I64u
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
这里有相当多的线段树讲解的实例：点击打开链接
题意：输入n，m；表示有n个数和m个询问。接着输入每个数的初始值。接着每个询问表示：Q表示求解i,j区间的和，C表示i~j区间的每个值都+k
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<algorithm>
#include<cstring>
#include<string>
#include<iostream>
const int MAXN=100000+10;
using namespace std;
typedef long long LL;
LL sum[MAXN<<2];
LL add[MAXN<<2];
LL ans;
struct node
{
int left,right;
}tree[MAXN<<2];
void down(int pos, int len)
{
if(add[pos])
{
add[pos<<1]+=add[pos];
add[pos<<1|1]+=add[pos];
sum[pos<<1]+=add[pos]*(len-(len>>1));//这里注意-和>>的优先级。。。
sum[pos<<1|1]+=add[pos]*(len>>1);
}
add[pos]=0;
}
void buildtree(int left, int right, int pos)
{
tree[pos].left=left;
tree[pos].right=right;
if(left==right){
scanf("%I64d", &sum[pos]);
return;
}
int mid=(left+right)>>1;
buildtree(left,mid,pos<<1);
buildtree(mid+1,right,pos<<1|1);
sum[pos]=sum[pos<<1]+sum[pos<<1|1];
}
void update(int pos, int l, int r, int val)
{
if(tree[pos].left==l && tree[pos].right==r){
add[pos]+=val;
sum[pos]+=val*(r-l+1);
return;
}
down(pos,tree[pos].right-tree[pos].left+1);
int mid=(tree[pos].left+tree[pos].right)>>1;
if(r<=mid) update(pos<<1,l,r,val);
else if(l>mid) update(pos<<1|1,l,r,val);
else{
update(pos<<1,l,mid,val);
update(pos<<1|1,mid+1,r,val);
}
sum[pos]=sum[pos<<1]+sum[pos<<1|1];
}
void query(int pos, int l, int r)
{
if(l<=tree[pos].left && tree[pos].right<=r){
ans+=sum[pos];
return;
}
down(pos,tree[pos].right-tree[pos].left+1);
int mid=(tree[pos].left+tree[pos].right)>>1;
if(r<=mid) query(pos<<1,l,r);
else if(l>mid) query(pos<<1|1,l,r);
else{
query(pos<<1,l,r);
query(pos<<1|1,l,r);
}
}
int main()
{
//freopen("in.txt","r",stdin);
int n,m;
while(scanf("%d%d", &n,&m)==2)
{
memset(add,0,sizeof(add));
int i,j,k;
buildtree(1,n,1);
char ch[3];
while(m--)
{
scanf("%s", ch);
if(ch[0]=='Q'){
scanf("%d%d", &i,&j);
ans=0;
query(1,i,j);
printf("%I64d\n", ans);
}
else{
scanf("%d%d%d", &i,&j,&k);
update(1,i,j,k);
}
}
}
return 0;
}