程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> poj3468 A Simple Problem with Integers（線段樹）

poj3468 A Simple Problem with Integers（線段樹）

編輯：C++入門知識

poj3468 A Simple Problem with Integers（線段樹）

A Simple Problem with Integers Time Limit: 5000MS Memory Limit: 131072K Total Submissions: 83200 Accepted: 25771 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.

Source

POJ Monthly--2007.11.25, Yang Yi
題意：一行序列，每次操作把一個區間裡的每個數都加上一個數，或者查詢一個區間的和。分析：一道模板題吧，就每次疊加就行了，只是注意數據范圍int會超。

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
#define ll long long
#define CL(a,b) memset(a,b,sizeof(a))
#define MAXN 100010

struct node
{
    int l,r;
    ll s,add;//add為每次加的數
}t[MAXN<<2];
int hh[MAXN];
int n,q;
ll ans;

void build(int l, int r, int i)
{
    t[i].l = l;
    t[i].r = r;
    t[i].add = 0;
    if(l == r) return ;
    int mid = (l+r)>>1;
    build(l, mid, i<<1);
    build(mid+1, r, i<<1|1);
    t[i].s = t[i<<1].s+t[i<<1|1].s;
}

void update(int l, int r, int add, int i)
{
    if(t[i].l>r || t[i].r=l && t[i].r<=r)
    {
        t[i].s += (t[i].r-t[i].l+1)*add;
        t[i].add += add;
        return ;
    }
    if(t[i].add)
    {
        t[i<<1].s += (t[i<<1].r-t[i<<1].l+1)*t[i].add;
        t[i<<1].add += t[i].add;
        t[i<<1|1].s += (t[i<<1|1].r-t[i<<1|1].l+1)*t[i].add;
        t[i<<1|1].add += t[i].add;
        t[i].add = 0;
    }
    update(l, r, add, i<<1);
    update(l, r, add, i<<1|1);
    t[i].s = t[i<<1].s+t[i<<1|1].s;
}

void query(int l, int r, int i)
{
    if(t[i].l>r || t[i].r=l && t[i].r<=r)
    {
        ans += t[i].s;
        return ;
    }
    if(t[i].add)
    {
        t[i<<1].s += (t[i<<1].r-t[i<<1].l+1)*t[i].add;
        t[i<<1].add += t[i].add;
        t[i<<1|1].s += (t[i<<1|1].r-t[i<<1|1].l+1)*t[i].add;
        t[i<<1|1].add += t[i].add;
        t[i].add = 0;
    }
    query(l, r, i<<1);
    query(l, r, i<<1|1);
    t[i].s = t[i<<1].s+t[i<<1|1].s;
}

int main()
{
    int a,b,c;
    ll k;
    char ch;
    while(scanf("%d%d",&n,&q)==2)
    {
        for(int i=1; i<=n; i++)
            scanf("%d",&hh[i]);
        build(1, n, 1);
        for(int i=1; i<=n; i++)
            update(i, i, hh[i], 1);
        while(q--)
        {
            getchar();
            scanf("%c",&ch);
            if(ch == 'C')
            {
                scanf("%d%d%d",&a,&b,&c);
                update(a, b, c, 1);
            }
            if(ch == 'Q')
            {
                ans = 0;
                scanf("%d%d",&a,&b);
                query(a, b, 1);
                printf("%lld\n",ans);
            }
        }
    }
    return 0;
}