E - A Simple Problem with Integers

最新推荐文章于 2025-05-27 22:22:47 发布

原创最新推荐文章于 2025-05-27 22:22:47 发布 · 537 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#解决方案 #I love ACM

简单数论专栏收录该内容

6 篇文章

订阅专栏

E - A Simple Problem with Integers

Time Limit:5000MS Memory Limit:131072KB 64bit IO Format:%I64d & %I64u

Submit Status

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 abc" means adding c to each of A_a, A_a₊₁, ... , A_b. -10000 ≤ c ≤ 10000.
"Q ab" 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

Hint

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

线段树功能:update:成段增减 query:区间求和

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
using namespace std;

#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1
#define LL long long
const int maxn = 111111;
LL add[maxn<<2];
LL sum[maxn<<2];
void PushUp(int rt)
{
    sum[rt] = sum[rt<<1] + sum[rt<<1|1];
}
void PushDown(int rt,int m)
{
    if (add[rt])
    {
        add[rt<<1] += add[rt];
        add[rt<<1|1] += add[rt];
        sum[rt<<1] += add[rt] * (m - (m >> 1));
        sum[rt<<1|1] += add[rt] * (m >> 1);
        add[rt] = 0;
    }
}
void build(int l,int r,int rt)
{
    add[rt] = 0;
    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 (m < R) 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;
}