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poj 3468 线段树成段更新

2024-07-12 03:30:32   231人阅读

http://poj.org/problem?id=3468

A Simple Problem with Integers
Time Limit: 5000MS Memory Limit: 131072K
Total Submissions: 58132 Accepted: 17704
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 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4

Sample Output

455915

Hint

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

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Life is so hard! many courage and not care the result!

 

成段更新不容易啊，看了两三天了，仍像云里雾里，不懂啊！

#include <stdio.h>#include <string.h>#include <stdlib.h>#include <algorithm>#define Maxx 100010int n,m;int str[Maxx];long long ans;struct Node{    int left,right;    long long add,sum;}tree[Maxx*4];void build(int l,int r,int rt){    tree[rt].left=l;    tree[rt].right=r;    tree[rt].add=0;        if(l == r)    {        tree[rt].sum=str[l];        return ;       }    int mid=(l+r)/2;    build(l,mid,2*rt);    build(mid+1,r,2*rt+1);    tree[rt].sum=tree[rt*2].sum+tree[rt*2+1].sum;}void update(int l,int r,int add,int rt){    if(tree[rt].left>=l && tree[rt].right<=r)    {        tree[rt].sum+=(tree[rt].right-tree[rt].left+1)*add;        tree[rt].add+=add;        return ;       }        if(tree[rt].left>r || tree[rt].right<l)    {        return ;       }    if(tree[rt].add)    {        tree[2*rt].sum += (tree[2*rt].right-tree[2*rt].left+1)*tree[rt].add;        tree[2*rt].add += tree[rt].add;        tree[2*rt+1].sum += (tree[2*rt+1].right-tree[2*rt+1].left+1)*tree[rt].add;        tree[2*rt+1].add += tree[rt].add;        tree[rt].add = 0;       }    update(l,r,add,rt*2);    update(l,r,add,rt*2+1);    tree[rt].sum =tree[rt*2].sum+tree[rt*2+1].sum;}void query(int l,int r,int rt){    if(tree[rt].left>r || tree[rt].right<l)    {            return ;       }    if(tree[rt].left>=l && tree[rt].right<=r)    {        ans+=tree[rt].sum;        return ;       }    if(tree[rt].add)    {        tree[2*rt].sum += (tree[2*rt].right-tree[2*rt].left+1)*tree[rt].add;        tree[2*rt].add += tree[rt].add;        tree[2*rt+1].sum += (tree[2*rt+1].right-tree[2*rt+1].left+1)*tree[rt].add;        tree[2*rt+1].add += tree[rt].add;        tree[rt].add = 0;       }    query(l,r,rt*2);    query(l,r,rt*2+1);    tree[rt].sum=tree[rt*2].sum+tree[rt*2+1].sum;}int main(){    int i,j,k,t,a,b,c;    while(scanf("%d%d",&n,&m)!=EOF)    {        for(i=1;i<=n;i++)        {            scanf("%d",&str[i]);           }        build(1,n,1);        char st[2];        for(i=1;i<=m;i++)        {                   scanf("%s",st);                 if(st[0] == ‘Q‘)                {                    scanf("%d%d",&a,&b);                    ans=0;                    query(a,b,1);                    printf("%lld\n",ans);                   }                else                {                    scanf("%d%d%d",&a,&b,&c);                    update(a,b,c,1);                   }           }           }    return 0;}

 


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