POJ-3468-A Simple Problem with Integers（线段树区间更新区间和）

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 139191		Accepted: 43086
Case Time Limit: 2000MS

Description

You have N integers, A1, A2, ... , AN. 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, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of Aa, Aa+1, ... , Ab.

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.

线段树模板题

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <queue>

#include <stack>

#define oo 0x3f3f3f3f

using namespace std;

struct node

{

    long long int lazy;

    long long int data;

    int l, r;

};

struct node tree[10000000];

long long int Begin[10000000];

void Buildtree( int root, int l, int r )

{

    tree[root].l = l;

    tree[root].r = r;

    tree[root].lazy = 0;

    if( l == r )

        tree[root].data = Begin[l];

    else

    {

        int mid = ( l + r ) >> 1;

        Buildtree( root<<1, l, mid );

        Buildtree( root<<1|1, mid+1, r);

        tree[root].data = tree[root<<1].data + tree[root<<1|1].data;

    }

}

void Pushdown( int root )

{

    if( tree[root].lazy != 0 )

    {

        tree[root<<1].lazy += tree[root].lazy;

        tree[root<<1|1].lazy += tree[root].lazy;

        tree[root<<1].data += ( tree[root<<1].r - tree[root<<1].l + 1 ) * tree[root].lazy;

        tree[root<<1|1].data += ( tree[root<<1|1].r - tree[root<<1|1].l + 1 ) * tree[root].lazy;

        tree[root].lazy = 0;

    }

}

void Updata( int root, int l, int r, int z )

{

    int i = tree[root].l, j = tree[root].r;

    if( i > r || l > j )

        return;

    if( i >= l && j <= r )

    {

        tree[root].data += (j - i + 1) * z;

        tree[root].lazy += z;

        return;

    }

    Pushdown( root );

    Updata( root<<1, l, r, z );

    Updata( root<<1|1, l, r, z );

    tree[root].data = tree[root<<1].data + tree[root<<1|1].data;

}

long long int Query ( int root, int l, int r )

{

    int i = tree[root].l, j = tree[root].r;

    if( i > r || l > j )

        return 0;

    if( l <= i && r >= j )

        return tree[root].data;

    Pushdown( root );

    return Query(root<<1, l, r) + Query(root<<1|1, l, r);

}

int main()

{

    int i, n, q;

    scanf("%d %d", &n, &q);

    for( i=1; i<=n; i++ )

        scanf("%lld", &Begin[i]);

    Buildtree( 1, 1, n );

    while( q-- )

    {

        char order;

        int a, b, c;

        getchar();

        scanf("%c", &order);

        if( order == 'C' )

        {

            scanf("%d %d %d", &a, &b, &c);

            Updata( 1, a, b, c);

        }

        else if( order == 'Q' )

        {

            scanf("%d %d", &a, &b);

            printf("%lld\n", Query( 1, a, b ));

        }

    }

    return 0;

}

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POJ-3468-A Simple Problem with Integers（线段树区间更新区间和）

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巴特西

POJ-3468-A Simple Problem with Integers（线段树 区间更新 区间和）

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POJ-3468-A Simple Problem with Integers（线段树区间更新区间和）