POJ A Simple Problem with Integers 线段树 lazy-target 区间跟新

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 105742		Accepted: 33031
Case Time Limit: 2000MS

Description

You have N integers, A₁, A₂, ... ,
A_N. 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, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.

Each of the next Q lines represents an operation.

"C a b c" means adding c to each of A_a,
A_a+1, ... , A_b. -10000 ≤ c ≤ 10000.

"Q a b" means querying the sum of A_a, A_a+1, ... ,
A_b.

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：线段树区间跟新 + lazy-target标记

（注意数据范围，long long）

代码：

#include<iostream>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
#include<cstdio>
using namespace std;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
const int maxn=100005;
long long sum[maxn<<2];
long long add[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];
        add[rt<<1|1]+=add[rt];
        sum[rt<<1]+=(len-(len>>1))*add[rt];
        sum[rt<<1|1]+=(len>>1)*add[rt];
        add[rt]=0;
    }
}
void build(int l, int r, int rt) {
    add[rt]=0;
    if(l==r) {
        scanf("%lld",&sum[rt]);
        return;
    }
    int mid=(l+r)>>1;
    build(lson);
    build(rson);
    pushup(rt);
}
void update(int L, int R, int val, int l, int r, int rt) {
    if(L<=l&&r<=R) {
        sum[rt]+=(r-l+1)*val;
        add[rt]+=val;
        return;
    }
    pushdown(rt,r-l+1);
    int mid=(l+r)>>1;
    if(L<=mid) update(L,R,val,lson);
    if(R>mid) update(L,R,val,rson);
    pushup(rt);
}
long long 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 mid=(l+r)>>1;
    long long cnt=0;
    if(L<=mid) cnt+=query(L,R,lson);
    if(R>mid) cnt+=query(L,R,rson);
    return cnt;
}
int main() {
    int n,q;
    while(~scanf("%d%d",&n,&q)) {
        build(1,n,1);
        char s[10];
        for(int i=1;i<=q;i++) {
            scanf("%s",s);
            if(s[0]=='Q') {
                int L,R;long long result;scanf("%d%d",&L,&R);
                result=query(L,R,1,n,1);
                printf("%lld\n",result);
            } else if(s[0]=='C') {
                int L,R,val;scanf("%d%d%d",&L,&R,&val);
                update(L,R,val,1,n,1);
            }
        }
    }
    return 0;
}