A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 163977		Accepted: 50540
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.

题意：　　Q查询区间和；C，将区间[x,y]的数都加上z

#include<iostream>

#include<algorithm>

#include<string.h>

#include<string>

#define ll long long

using namespace std;

ll tree[], lazy[], len[];//tree[num]存的是节点num所在区间的区间和

void pushdown(ll num)

{

    if (lazy[num] != )

    {

        tree[num * ] = tree[num * ] + lazy[num] * len[num * ];

        tree[num *  + ] = tree[num *  + ] + lazy[num] * len[num *  + ];

        lazy[num * ] = lazy[num * ] + lazy[num];

        lazy[num *  + ] = lazy[num *  + ] + lazy[num];

        lazy[num] = ;

    }

}

void build(ll num, ll le, ll ri)

{

    len[num] = ri - le + ;//区间长度

    if (le == ri)

    {

        scanf("%lld", &tree[num]);

        return;

    }

    ll mid = (le + ri) / ;

    build(num * , le, mid);

    build(num *  + , mid + , ri);

    tree[num] = tree[num * ] + tree[num *  + ];

}

void update(ll num, ll le, ll ri, ll x, ll y, ll z)

{

    if (x <= le && ri <= y)

    {

        lazy[num] = lazy[num] + z;

        tree[num] = tree[num] + len[num] * z;//更新区间和

        return;

    }

    pushdown(num);

    ll mid = (le + ri) / ;

    if (x <= mid)

        update(num * , le, mid, x, y, z);

    if (y > mid)

        update(num *  + , mid + , ri, x, y, z);

    tree[num]=tree[num*]+tree[num*+];

}

ll query(ll num, ll le, ll ri, ll x, ll y)

{

    if (x <= le && ri <= y)//查询区间在num节点所在区间内

        return tree[num];

    pushdown(num);

    ll mid = (le + ri) / ;

    ll ans = ;

    if (x <= mid)

        ans = ans + query(num * , le, mid, x, y);

    if (y > mid)

        ans = ans + query(num *  + , mid + , ri, x, y);

    return ans;

}

int main()

{

    ll n, m;

    scanf("%lld%lld", &n, &m);

    build(, , n);

    while (m--)

    {

        char c[];

        scanf("%s", c);

        if (c[] == 'Q')

        {

            ll x, y;

            scanf("%lld%lld", &x, &y);

            printf("%lld\n", query(, , n, x, y));

        }

        else

        {

            ll x, y, z;

            scanf("%lld%lld%lld", &x, &y, &z);

            update(, , n, x, y, z);

        }

    }

    return ;

}

巴特西

POJ 3468 区间更新（求任意区间和）A Simple Problem with Integers

A Simple Problem with Integers

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