• poj 3468 A Simple Problem with Integers【线段树区间修改】


    A Simple Problem with Integers
    Time Limit: 5000MS   Memory Limit: 131072K
    Total Submissions: 79137   Accepted: 24395
    Case Time Limit: 2000MS

    Description

    You have N integers, A1A2, ... , 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 A1A2, ... , 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.
     
    #include<stdio.h>
    #include<string.h>
    #include<algorithm>
    #define MAX 100000+10 
    using namespace std;
    long long add[MAX<<2];
    long long sum[MAX<<2];
    void pushup(int o)
    {
    	sum[o]=sum[o<<1]+sum[o<<1|1];
    }
    void pushdown(int o,int m)
    {
    	if(add[o])
    	{
    		add[o<<1]+=add[o];
    		add[o<<1|1]+=add[o];
    		sum[o<<1]+=add[o]*(m-(m>>1));
    		sum[o<<1|1]+=add[o]*(m>>1);
    		add[o]=0;
    	}
    }
    void gettree(int o,int l,int r)
    {
    	add[o]=0;
    	if(l==r)
    	{
    		scanf("%lld",&sum[o]);
    		return ;
    	}
    	int mid=(l+r)>>1;
    	gettree(o<<1,l,mid);
    	gettree(o<<1|1,mid+1,r);
    	pushup(o);
    }
    void update(int o,int l,int r,int L,int R,int val)
    {
    	if(L<=l&&R>=r)
    	{
    		add[o]+=val;
    		sum[o]+=val*(r-l+1);
    		return ;
    	}
    	pushdown(o,r-l+1);
    	int mid=(l+r)>>1;
    	if(L<=mid)
    	    update(o<<1,l,mid,L,R,val);
    	if(R>mid)
    	    update(o<<1|1,mid+1,r,L,R,val);
    	pushup(o);
    }
    long long find(int o,int l,int r,int L,int R)
    {
    	if(L<=l&&R>=r)
    	{
    		return sum[o];
    	}
    	pushdown(o,r-l+1);
    	long long ans=0;
    	int mid=(l+r)>>1;
    	if(L<=mid)
    	    ans+=find(o<<1,l,mid,L,R);
    	if(R>mid)
    	    ans+=find(o<<1|1,mid+1,r,L,R);
    	return ans; 
    }
    int main()
    {
    	int n,m,i,j;
    	int a,b,c;
    	char d[12];
    	while(scanf("%d%d",&n,&m)!=EOF)
    	{
    		gettree(1,1,n);
    		for(i=1;i<=m;i++)
    		{
    			scanf("%s%d%d",d,&a,&b);
    			if(d[0]=='Q')
    			    printf("%lld
    ",find(1,1,n,a,b));
    			else
    			{
    				scanf("%d",&c);
    				update(1,1,n,a,b,c);
    			}			    
    		}
    	} 
    	return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/tonghao/p/4792865.html
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