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|1
using 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;
}




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