kuangbin专题七 POJ3468 A Simple Problem with Integers （线段树或树状数组）

You have N integers, A₁, A₂, ... , A_N. 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, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a+1, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a, A_a+1, ... , A_b.

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

区间修改 区间查询  注意pushdown的操作就可以了

 #include <iostream>
 #include <stdio.h>
 #include <math.h>
 #include <string.h>
 #include <stdlib.h>
 #include <string>
 #include <vector>
 #include <set>
 #include <map>
 #include <queue>
 #include <algorithm>
 #include <sstream>
 #include <stack>
 using namespace std;
 #define FO freopen("in.txt","r",stdin);
 #define rep(i,a,n) for (int i=a;i<n;i++)
 #define per(i,a,n) for (int i=n-1;i>=a;i--)
 #define pb push_back
 #define mp make_pair
 #define all(x) (x).begin(),(x).end()
 #define fi first
 #define se second
 #define SZ(x) ((int)(x).size())
 #define debug(x) cout << "&&" << x << "&&" << endl;
 #define lowbit(x) (x&-x)
 #define mem(a,b) memset(a, b, sizeof(a));
 typedef vector<int> VI;
 typedef long long ll;
 typedef pair<int,int> PII;
 const ll mod=;
 const int inf = 0x3f3f3f3f;
 ll powmod(ll a,ll b) {ll res=;a%=mod;for(;b;b>>=){if(b&)res=res*a%mod;a=a*a%mod;}return res;}
 ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
 //head

 //区间修改 区间查询
 const int maxx=;
 ll sum[maxx<<],lazy[maxx<<],val;
 int n,q;

 void Pushup(int rt) {
     sum[rt]=sum[rt<<]+sum[rt<<|];
 }

 void build(int rt,int L,int R) {
     lazy[rt]=;
     if(L==R) {
         scanf("%lld",&sum[rt]);
         return;
     }
     int mid=(L+R)>>;
     build(rt<<,L,mid);
     build(rt<<|,mid+,R);
     Pushup(rt);
 }

 void Pushdown(int rt,int x) {
     if(lazy[rt]) {
         lazy[rt<<]+=lazy[rt];
         lazy[rt<<|]+=lazy[rt];
         sum[rt<<]+=(x-(x>>))*lazy[rt];//左子树加左边一半的值
         sum[rt<<|]+=(x>>)*lazy[rt];//右子树同理
         lazy[rt]=;//清空
     }
 }

 void Updata(int rt,int L,int R,int l,int r) {
     if(L>=l&&R<=r) {
         lazy[rt]+=val;//累加标记
         sum[rt]+=(R-L+)*val;//更新值
         return;
     }
     int mid=(L+R)>>;
     Pushdown(rt,R-L+);//这里多了一个参数 L R区间的个数
     if(l<=mid) Updata(rt<<,L,mid,l,r);
     if(r>mid) Updata(rt<<|,mid+,R,l,r);
     Pushup(rt);
 }

 ll Query(int rt,int L,int R,int l,int r) {
     if(L>=l&&R<=r)
         return sum[rt];
     ll ans=;
     int mid=(L+R)>>;
     Pushdown(rt,R-L+);
     if(l<=mid) ans+=Query(rt<<,L,mid,l,r);
     if(r>mid) ans+=Query(rt<<|,mid+,R,l,r);
     Pushup(rt);
     return ans;
 }

 int main() {
     while(~scanf("%d%d",&n,&q)) {
         build(,,n);
         char s[];
         while(q--) {
             scanf("%s",s);
             int l,r;
             if(s[]=='Q') {
                 scanf("%d%d",&l,&r);
                 printf("%lld\n",Query(,,n,l,r));
             } else {
                 scanf("%d%d%lld",&l,&r,&val);
                 Updata(,,n,l,r);
             }
         }
     }
 }

借此机会学习了一波树状数组，有点难理解。

单点修改 区间查询  Updata(l,val)

区间修改 单点查询  引入差分数组 c[i]=d[i]-d[i-1]; Updata(l,val);  Updata(r+1,-val); ----明白树状数组就好理解了。把多加的减去

区间修改 区间查询  公式可以推到一下，sum[n]=n*(c[1]+c[2]+...c[n])-(0*c[1]+1*c[2]+2*c[3]+...+(n-1)*c[n]);

所以再维护一个 (i-1)*(a[i]-a[i-1]) , a为原数组。

Updata(l,(l-1)*val); Updata(r+1,-r*val);   原理同上。多加的减去。

（纯属个人理解。关键还是公式的推导，和树状数组的理解）

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 typedef long long ll;
 #define lowbit(x) (x&-x)

 int n,q;
 const int maxn=;
 ll a[maxn],val,c1[maxn],c2[maxn];//维护两个树状数组

 void Updata(ll c[],int pos,ll val) {
     while(pos<=n) {
         c[pos]+=val;
         pos+=lowbit(pos);
     }
 } 

 ll getsum(ll c[],int pos) {
     ll ans=;
     while(pos>) {
         ans+=c[pos];
         pos-=lowbit(pos);
     }
     return ans;
 }

 ll sigma(int r) {//公式推导
     ll sum1=r*getsum(c1,r);
     ll sum2=getsum(c2,r);
     return sum1-sum2;
 }

 ll Query(int l,int r) {//区间求和
     return sigma(r)-sigma(l-);
 }

 int main() {
     while(~scanf("%d%d",&n,&q)) {
         memset(c1,,sizeof(c1));memset(c2,,sizeof(c2));
         for(int i=;i<=n;i++) {
             scanf("%lld",&a[i]);
             Updata(c1,i,a[i]-a[i-]);//维护a[i]-a[i-1]
             Updata(c2,i,(i-)*(a[i]-a[i-]));//维护(i-1)*(a[i]-a[i-1])
         }
         char s[];
         while(q--) {
             int l,r;
             scanf("%s",s);
             if(s[]=='Q') {
                 scanf("%d%d",&l,&r);
                 printf("%lld\n",Query(l,r));
             } else {
                 scanf("%d%d%lld",&l,&r,&val);
                 //更新操作有点难理解
                 Updata(c1,l,val);Updata(c1,r+,-val);
                 Updata(c2,l,(l-)*val);Updata(c2,r+,-r*val);
             }
         }
     }
 }