POJ3468 A Simple Problem with Integers

一.原题链接： http://poj.org/problem?id=3468
二.题目大意：给出1到N个数，2种操作，Q a b查询从第a个数到第b个数的总和，C a b c从第a个数到第b个数每个数加上c。对于每次查询，输出总和。
三.思路：线段树成段更新模板题。所谓成段更新，就是加一个lazy标记，缓存要增加的元素，每次更新不更新到底，而是累积到lazy数组中，原题为add数组。要进行更深的查询或者更深的更新，用pushDown向下更新，这样就可以避免不必要的更新。
四.代码：
#include <cstdio>
#include <cmath>

using namespace std;

const int MAX_N = 100100;
const int INF = 0x3f3f3f3f;

#define LC(t) t<<1
#define RC(t) t<<1|1

struct node
{
int l, r;
int mid()
{
return l+r >> 1;
}
}seTree[4*MAX_N];

long long add[MAX_N<<2],sum[MAX_N<<2];

void build(int nd, int l, int r)
{
seTree[nd].l = l;
seTree[nd].r = r;
sum[nd] = add[nd] = 0;
if(l == r)
return;
int mid = seTree[nd].mid();
build(LC(nd), l, mid);
build(RC(nd), mid+1, r);
}

void pushDown(int nd, int len)
{
if(!add[nd])
return;
add[LC(nd)] += add[nd];
add[RC(nd)] += add[nd];
sum[LC(nd)] += add[nd]*(len -(len>>1));
sum[RC(nd)] += add[nd]*(len>>1);
add[nd] = 0;
}

void pushUp(int nd)
{
sum[nd] = sum[LC(nd)] + sum[RC(nd)];
}

void upDate(int nd, int l, int r,int ele)
{
if(l == seTree[nd].l && r ==seTree[nd].r){
sum[nd] += ele*(r-l+1);
add[nd] += ele;
return;
}

if(seTree[nd].l == seTree[nd].r)
return;
pushDown(nd, seTree[nd].r-seTree[nd].l+1);
int mid = seTree[nd].mid();

if(r <= mid)
upDate(LC(nd), l, r, ele);
else if(l > mid)
upDate(RC(nd), l, r, ele);
else{
upDate(LC(nd), l, mid, ele);
upDate(RC(nd), mid+1, r, ele);
}
pushUp(nd);
}

long long query(int nd, int l,int r)
{
if(seTree[nd].l == l &&seTree[nd].r == r)
return sum[nd];

pushDown(nd, seTree[nd].r-seTree[nd].l+1);
int mid = seTree[nd].mid();

if(r <= mid)
return query(LC(nd), l, r);
if(l > mid)
return query(RC(nd), l, r);
return query(LC(nd), l, mid) +query(RC(nd), mid+1, r);
}

int main()
{
//freopen("in.txt","r", stdin);

int N, Q, i, ele, a, b, c;
char ch;

while(~scanf("%d%d", &N,&Q)){
build(1, 1, N);

for(i = 1; i <= N; i++){
scanf("%d", &ele);
upDate(1, i, i, ele);
}

for(i = 1; i <= Q; i++){
scanf("\n%c %d %d",&ch, &a, &b);
if(ch == 'C'){
scanf("%d", &c);
upDate(1, a, b, c);
}
else
printf("%I64d\n",query(1, a, b));
}
}
return 0;
}


五. 易错点：本渣还是弄了好久，一开始混淆了线段树的左右边界和要查询和更新的左右边界，然后还犯了奇数整除会损失精度的错误，也就是说len-len/2与len/2是不一样的。