POJ---3468---A Simple Problem with Integers

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 94448		Accepted: 29421
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.

Source

POJ Monthly--2007.11.25, Yang Yi

 

POJ---3264（同类型的吧）

思路：

线段树成段更新的入门题目。。学会使用lazy即可。还需要注意的是，lazy的时候更改是累加，而不是直接修改。。有可能连续几次进行修改操作。。注意这一点就好了。。

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
#include <math.h>
#include <vector>
using namespace std;
typedef long long LL;
const int oo = 0xfffffff;
const int maxn = 110007;
#define lson rt<<1
#define rson rt<<1|1
int A[maxn];

struct da
{
    int l, r;
    int len()
    {
        return r-l+1;
    }
    LL s, e;
} tree[maxn<<2];

void build(int l, int r, int rt)// 建立一颗树
{
    tree[rt].l = l;
    tree[rt].r = r;
    tree[rt].e = 0;
    if(l == r)
    {
        tree[rt].s = A[l];
        return ;
    }
    int mid = (l+r)/2;
    build(l, mid, rt*2);
    build(mid+1, r, rson);
    tree[rt].s = tree[lson].s + tree[rson].s;
}

void down(int rt)
{
    tree[lson].s += tree[lson].len()*tree[rt].e;
    tree[rson].s += tree[rson].len()*tree[rt].e;

    tree[lson].e += tree[rt].e;
    tree[rson].e += tree[rt].e;

    tree[rt].e = 0;
}
void update(int l, int r, int val, int rt)
{
    tree[rt].s += val*(r-l+1);
    if(tree[rt].l == l && r == tree[rt].r)///超时的原因****&*
    {
        tree[rt].e += val;
        return;
    }

    down(rt);

    int mid = (tree[rt].l+tree[rt].r)/2;
    if(r <= mid) update(l, r, val, lson);
    else if(l > mid)update(l, r, val, rson);
    else
    {
        update(l, mid, val, lson);
        update(mid+1, r, val,rson);
    }
}
LL query(int l, int r, int rt)
{
    if(tree[rt].l == l && r == tree[rt].r)
    {
        return tree[rt].s;
    }

    down(rt);

    int mid = (tree[rt].l+tree[rt].r)/2;
    if(r <= mid)
        return query(l, r, lson);
    else if(l > mid)
        return query(l, r, rson);
    else
    {
        LL lsum, rsum;
        lsum = query(l, mid, lson);
        rsum = query(mid+1, r, rson);
        return lsum+rsum ;
    }
}
int main()
{
    int n, q, i, a, b, c;
    while(scanf("%d %d", &n, &q) != EOF)
    {
        char op[20];
        for(i = 1; i <= n; i++)
            scanf("%d", &A[i]);

        build(1, n, 1);
        while(q--)
        {
            scanf("%s %d %d", op, &a, &b);
            if(op[0] == 'Q')
            {
                 printf("%lld
", query(a, b, 1));
                 continue;
            }
            scanf("%d", &c);
            update(a, b, c, 1);
        }
    }
    return 0;
}