POJ3468_A Simple Problem with Integers(线段树/成段更新)

解题报告

题意：

略

思路：

线段树成段更新，区间求和。

#include <iostream>
#include <cstring>
#include <cstdio>
#define LL long long
#define int_now int l,int r,int root
using namespace std;
LL sum[500000],lazy[500000];
void push_up(int root,int l,int r) {
    sum[root]=sum[root*2]+sum[root*2+1];
}
void push_down(int rt,int l,int r) {
    if(lazy[rt]) {
        int m=(r-l+1);
        lazy[rt<<1]+=lazy[rt];
        lazy[rt<<1|1]+=lazy[rt];
        sum[rt<<1]+=lazy[rt]*(m-(m/2));
        sum[rt<<1|1]+=lazy[rt]*(m/2);
        lazy[rt]=0;
    }
}
void update(int root,int l,int r,int ql,int qr,LL v) {
    if(ql>r||qr<l)return;
    if(ql<=l&&r<=qr) {
        lazy[root]+=v;
        sum[root]+=v*(r-l+1);
        return ;
    }
    int mid=(l+r)/2;
    push_down(root,l,r);
    update(root*2,l,mid,ql,qr,v);
    update(root*2+1,mid+1,r,ql,qr,v);
    push_up(root,l,r);
}
LL q_sum(int root,int l,int r,int ql,int qr) {
    if(ql>r||qr<l)return 0;
    if(ql<=l&&r<=qr)return sum[root];
    push_down(root,l,r);
    int mid=(l+r)/2;
    return q_sum(root*2,l,mid,ql,qr)+q_sum(root*2+1,mid+1,r,ql,qr);
}
int main() {
    int n,q,i,j,ql,qr;
    LL a;
    scanf("%d%d",&n,&q);
    for(i=1; i<=n; i++) {
        scanf("%lld",&a);
        update(1,1,n,i,i,a);
    }
    char str[10];
    for(i=1; i<=q; i++) {
        scanf("%s",str);
        if(str[0]=='Q') {
            scanf("%d%d",&ql,&qr);
            printf("%lld
",q_sum(1,1,n,ql,qr));
        } else {
            scanf("%d%d%lld",&ql,&qr,&a);
            update(1,1,n,ql,qr,a);
        }
    }
    return 0;
}

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 60817		Accepted: 18545
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

Hint

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