POJ 3468（线段树，区间加减 询问区间和）

problem

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

Hint

The sums may exceed the range of 32-bit integers.

思路

裸的模板题

代码示例

#include<iostream>#include<cstdio>#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1using namespace std;const int maxn=200010;typedef long long LL;LL add[maxn<<2];//兼顾lazy标记与具体变化LL sum[maxn<<2];void PushUp(int rt){    sum[rt]=sum[rt<<1]+sum[rt<<1|1];}void PushDown(int rt,int len){    if(add[rt]){        add[rt<<1]+=add[rt];//下放，注意是+=而不是直接赋值        add[rt<<1|1]+=add[rt];        sum[rt<<1]+=add[rt]*(len-(len>>1));        sum[rt<<1|1]+=add[rt]*(len>>1);        add[rt]=0;//下放了，清除标记    }}void build(int l,int r,int rt){    add[rt]=0;//初始化    if(l==r){        scanf("%lld",&sum[rt]);        return ;    }    int m=(l+r)>>1;    build(lson); build(rson);    PushUp(rt);}void update(int L,int R,int c,int l,int r,int rt){    if(L<=l&&r<=R){        add[rt]+=c;        sum[rt]+=(LL)c*(r-l+1);        return ;    }    PushDown(rt,r-l+1);    int m=(l+r)>>1;    if(L<=m) update(L,R,c,lson);    if(m<R) update(L,R,c,rson);    PushUp(rt);}LL query(int L,int R,int l,int r,int rt){    if(L<=l&&r<=R)  return sum[rt];    PushDown(rt,r-l+1);    int m=(l+r)>>1;    LL ret=0;    if(L<=m) ret+=query(L,R,lson);    if(R>m) ret+=query(L,R,rson);    return ret;}int main(){    int N,Q;    scanf("%d%d",&N,&Q);    build(1,N,1);    while(Q--)    {        char op[2];        int a,b,c;        scanf("%s",op);        if(op[0]=='Q'){            scanf("%d%d",&a,&b);            printf("%lld\n",query(a,b,1,N,1));        }        else{            scanf("%d%d%d",&a,&b,&c);            update(a,b,c,1,N,1);        }    }    return 0;}