poj 3468 A Simple Problem with Integers LAZY线段树

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A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072K

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

思路：基础的懒散线段树练习   (水完这道就不水了)

代码：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<string>#include<vector>#include <ctime>#include<queue>#include<set>#include<map>#include<stack>#include<iomanip>#include<cmath>#define mst(ss,b) memset((ss),(b),sizeof(ss))#define maxn 0x3f3f3f3f#define MAX 1000100///#pragma comment(linker, "/STACK:102400000,102400000")typedef long long ll;typedef unsigned long long ull;#define INF (1ll<<60)-1using namespace std;int n,q;ll a[400100],sum[400100],lazy[400100];char s[10];void build(int root,int l,int r){    if(l==r){        sum[root]=a[l];        return ;    }    int mid=(l+r)/2;    build(root*2,l,mid);    build(root*2+1,mid+1,r);    sum[root]=sum[root*2]+sum[root*2+1];}void pushdown(int root,int l){    if(lazy[root]){        lazy[root*2]+=lazy[root];        lazy[root*2+1]+=lazy[root];        sum[root*2]+=(ll)(l-l/2)*lazy[root];        sum[root*2+1]+=(ll)(l/2)*lazy[root];        lazy[root]=0;    }}void update(int root,int l,int r,int x,int y,ll z){    if(l>=x && r<=y){        sum[root]+=(ll)(r-l+1)*z;        lazy[root]+=z;        return ;    }    pushdown(root,r-l+1);    int mid=(l+r)/2;    if(y<=mid) update(root*2,l,mid,x,y,z);    else if(x>mid) update(root*2+1,mid+1,r,x,y,z);    else {        update(root*2,l,mid,x,mid,z);        update(root*2+1,mid+1,r,mid+1,y,z);    }    sum[root]=sum[root*2]+sum[root*2+1];}ll query(int root,int l,int r,int x,int y){    if(l>=x && r<=y){        ///cout<<"root="<<root<<" "<<sum[root]<<endl;        return sum[root];    }    pushdown(root,r-l+1);    int mid=(l+r)/2;    if(y<=mid) return query(root*2,l,mid,x,y);    else if(x>mid) return query(root*2+1,mid+1,r,x,y);    else {        return query(root*2,l,mid,x,mid)+query(root*2+1,mid+1,r,mid+1,y);    }}int main(){    while(scanf("%d%d",&n,&q)!=EOF){        mst(sum,0);        mst(lazy,0);        for(int i=1;i<=n;i++) scanf("%lld",&a[i]);        build(1,1,n);        while(q--){            scanf("%s",s);            int x,y;            ll z;            if(s[0]=='Q'){                scanf("%d%d",&x,&y);                printf("%lld\n",query(1,1,n,x,y));            } else {                scanf("%d%d%lld",&x,&y,&z);                update(1,1,n,x,y,z);            }        }    }    return 0;}<strong></strong>