ACM刷题之POJ————A Simple Problem with Integers

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#ACM #线段树 #poj

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本文介绍了一种基于线段树的数据结构解决区间加法操作及区间求和查询的问题。通过具体的AC代码实现，展示了如何高效地处理大规模数据集中的这类操作。

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A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 106659		Accepted: 33284
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

Hint

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

线段树的裸题

不过这里多了一个区间更新，要注意一下

下面是ac代码

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<algorithm>
#include<map>
#include<set>
#include<queue>
#include<string>
#include<iostream>
using namespace std;
#define MID(x,y) ((x+y)>>1)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int N=2e5+7;

#define max(a,b) (a>b)?a:b
#define min(a,b) (a>b)?b:a
#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1
#define LL __int64
const int maxn = 100100;
using namespace std;

LL lazy[maxn<<2];
LL sum[maxn<<2];

void putup(int rt)
{
  sum[rt] = sum[rt<<1] + sum[rt<<1|1];
}

void putdown(int rt,int m)
{
  if (lazy[rt])
  {
    lazy[rt<<1] += lazy[rt];
    lazy[rt<<1|1] += lazy[rt];
    sum[rt<<1] += lazy[rt] * (m - (m >> 1));
    sum[rt<<1|1] += lazy[rt] * (m >> 1);
    lazy[rt] = 0;
  }
}

void build(int l,int r,int rt) {
  lazy[rt] = 0;
  if (l == r)
  {
    scanf("%I64d",&sum[rt]);
    return ;
  }
  int m = (l + r) >> 1;
  build(lson);
  build(rson);
  putup(rt);
}

void update(int L,int R,int c,int l,int r,int rt)
{
  if (L <= l && r <= R)
  {
    lazy[rt] += c;
    sum[rt] += (LL)c * (r - l + 1);
    return ;
  }
  putdown(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);
  putup(rt);
}

LL query(int L,int R,int l,int r,int rt)
{
  if (L <= l && r <= R)
  {
    return sum[rt];
  }
  putdown(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 , m;int a , b , c;
  char str[5];
  scanf("%d%d",&n,&m);
  build(1 , n , 1);
  while (m--)
  {
    scanf("%s",str);
    if (str[0] == 'Q')
    {
      scanf("%d%d",&a,&b);
      printf("%I64d\n",query(a , b , 1 , n , 1));
    }
    else if(str[0]=='C')
    {
      scanf("%d%d%d",&a,&b,&c);
      update(a , b , c , 1 , n , 1);
    }
  }
  return 0;
}