poj3468 A Simple Problem with Integers （线段树的懒惰标记）

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题目链接：

http://poj.org/problem?id=3468

题意：

对于一个区间的数有两种操作：

C a b c 在区间[a,b]内的数全部加上c

Q a b 查询区间[a,b]之间的数的和

解：

嗯，没什么好说的，就直接线段树就行了，需要注意的就是懒惰标记的理解和应用了。

关于懒惰标记的理解推荐这篇博客：http://blog.csdn.net/sdjzping/article/details/19542103

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <algorithm>using namespace std;const int maxn=1e5+50;typedef long long ll;ll a[maxn];int n;struct Tree{int node,l,r;ll v,laz;}seg[ maxn << 2 ];void pushup(int node){seg[node].v=seg[node<<1].v+seg[node<<1|1].v;}void pushdown(int node,int b){    if(seg[node].laz)    {        seg[node<<1].laz+=seg[node].laz;        seg[node<<1|1].laz+=seg[node].laz;        seg[node<<1].v+=(ll)(b-(b/2))*seg[node].laz;        seg[node<<1|1].v+=(ll)(b/2)*seg[node].laz;        seg[node].laz=0;    }}void build(int l,int r,int node){    seg[node].l=l;seg[node].r=r;seg[node].laz=0;    if(l==r){    seg[node].v=a[l];    return ;    }    int mid=(l+r)/2;    build(l,mid,node<<1);    build(mid+1,r,node<<1|1);    pushup(node);}void update(int l,int r,int node,int k){    if(l<=seg[node].l&&seg[node].r<=r)    {        seg[node].v+=(seg[node].r-seg[node].l+1)*k;        seg[node].laz+=k;        return ;    }    pushdown(node,seg[node].r-seg[node].l+1);    int mid=(seg[node].l+seg[node].r)/2;    if(mid>=l)    update(l,r,node<<1,k);    if(r>mid)    update(l,r,node<<1|1,k);    pushup(node);}ll query(int l,int r,int node){    if(l<=seg[node].l&&seg[node].r<=r)    {        //cout<<"Node:"<<node<<" L:"<<seg[node].l<<" R:"<<seg[node].r<<endl;        return seg[node].v;    }    pushdown(node,seg[node].r-seg[node].l+1);    int mid=(seg[node].l+seg[node].r)/2;    ll  xa=0,xb=0;    if(mid>=l)    xa=query(l,r,node<<1);    if(r>mid)    xb=query(l,r,node<<1|1);    pushup(node);    //cout<<"Node:"<<node<<" Xa:"<<xa<<" Xb:"<<xb<<endl;    return xa+xb;}void show(int node){    cout<<"Node:"<<node<<endl;    cout<<"L:"<<seg[node].l<<" R:"<<seg[node].r<<endl;    cout<<"Value:"<<seg[node].v<<endl;    cout<<endl;    if(seg[node].l==seg[node].r)return ;    show(node<<1);    show(node<<1|1);}int main(){    int n,q;    while(~scanf("%d%d",&n,&q))    {        for(int i=1;i<=n;i++)        scanf("%I64d",&a[i]);        build(1,n,1);        while(q--){        char op[2];        int c1,c2;        scanf("%s",op);        if(op[0]=='C')        {            int c3;            scanf("%d%d%d",&c1,&c2,&c3);            update(c1,c2,1,c3);            //show(1);        }        else        {            scanf("%d%d",&c1,&c2);            ll ans=query(c1,c2,1);            printf("%I64d\n",ans);        }        }    }    return 0;}

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