• POJ 3468:A Simple Problem with Integers(线段树区间更新模板)


                            A Simple Problem with Integers

    Time Limit: 5000MS   Memory Limit: 131072K
    Total Submissions: 141093   Accepted: 43762
    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 AaAa+1, ... , Ab. -10000 ≤ c ≤ 10000.
    "Q a b" means querying the sum of AaAa+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

    4
    55
    9
    15

    Hint

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

    题意

    线段树区间更新求和模板题。

    给出n个数和m次操作,Q表示查询该区间内的数的和并输出,C表示把该区间内的所有数都加上一个值。

    AC代码

    抄的师傅的板子,还不是太理解

    #include <stdio.h>
    #include <string.h>
    #include <iostream>
    #include <algorithm>
    #include <math.h>
    #include <limits.h>
    #include <map>
    #include <stack>
    #include <queue>
    #include <vector>
    #include <set>
    #include <string>
    #define ll long long
    #define ull unsigned long long
    #define ms(a) memset(a,0,sizeof(a))
    #define pi acos(-1.0)
    #define INF 0x7f7f7f7f
    #define lson o<<1
    #define rson o<<1|1
    const double E=exp(1);
    const int maxn=2e5+10;
    const int mod=1e9+7;
    using namespace std;
    struct wzy
    {
    	ll left,right,len;
    	ll value;
    	ll lazy;
    }p[maxn<<2];
    void push_up(ll o)
    {
    	p[o].value=p[lson].value+p[rson].value;
    }
    void push_down(ll o)
    {
    	if(p[o].lazy)
    	{
    		p[lson].lazy+=p[o].lazy;
    		p[rson].lazy+=p[o].lazy;
    		p[lson].value+=p[o].lazy*p[lson].len;
    		p[rson].value+=p[o].lazy*p[rson].len;
    		p[o].lazy=0;
    	}
    }
    void build(ll o,ll l,ll r)
    {
    	p[o].left=l;p[o].right=r;
    	p[o].len=r-l+1;
    	p[o].lazy=0;
    	if(l==r)
    	{
    		ll x;
    		scanf("%lld",&x);
    		p[o].value=x;
    		return ;
    	}
    	ll mid=(l+r)>>1;
    	build(lson,l,mid);
    	build(rson,mid+1,r);
    	push_up(o);
    }
    void update(ll o,ll l,ll r,ll v)
    {
    	if(p[o].left>=l&&p[o].right<=r)
    	{
    		p[o].lazy+=v;
    		p[o].value+=v*p[o].len;
    		return ;	
    	}
    	push_down(o);
    	ll mid=(p[o].left+p[o].right)>>1;
    	if(r<=mid)
    		update(lson,l,r,v);
    	else if(l>mid)
    		update(rson,l,r,v);
    	else
    	{
    		update(lson,l,mid,v);
    		update(rson,mid+1,r,v);
    	}
    	push_up(o);
    }
    ll query(ll o,ll l,ll r)
    {
    	if(p[o].left>=l&&p[o].right<=r)
    		return p[o].value;
    	ll mid=(p[o].left+p[o].right)>>1;
    	push_down(o);
    	if(r<=mid)
    		return query(lson,l,r);
    	else if(l>mid)
    		return query(rson,l,r);
    	else
    		return query(lson,l,mid)+query(rson,mid+1,r);
    }
    int main(int argc, char const *argv[])
    {
    	int n,m;
    	scanf("%d%d",&n,&m);
    	build(1,1,n);
    	char ch[5];
    	ll a,b,c;
    	while(m--)
    	{
    		scanf("%s",ch);
    		if(ch[0]=='Q')
    		{
    			scanf("%lld%lld",&a,&b);
    			printf("%lld
    ",query(1,a,b));
    		}
    		else
    		{
    			scanf("%lld%lld%lld",&a,&b,&c);
    			update(1,a,b,c);
    		}
    	}
    	return 0;
    }
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  • 原文地址:https://www.cnblogs.com/Friends-A/p/10324380.html
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