• poj 3468 A Simple Problem with Integers(线段树区间更新)


    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


     1 #include <iostream>
     2 #include <cstdio>
     3 using namespace std;
     4 #define LL long long
     5 #define lson rt<<1,first,mid
     6 #define rson rt<<1 | 1,mid+1,end
     7 const int maxn=100005;
     8 LL tree[maxn<<2];
     9 LL add[maxn<<2];
    10 void push_up(int rt)
    11 {
    12     tree[rt]=tree[rt<<1]+tree[rt<<1 | 1];
    13 }
    14 void push_down(int rt,int m) //更新子节点的数值
    15 {
    16     if(add[rt])
    17     {
    18         add[rt<<1]+=add[rt];
    19         add[rt<<1|1]+=add[rt];
    20         tree[rt<<1]+=add[rt]*(m-(m>>1));
    21         tree[rt<<1|1]+=add[rt]*(m>>1);
    22         add[rt]=0; //更新后需要还原
    23     }
    24 }
    25 void build(int rt,int first,int end)
    26 {
    27     add[rt]=0;
    28     if(first==end)
    29     {
    30         scanf("%lld",&tree[rt]);
    31         //cin>>tree[rt];
    32         return ;
    33     }
    34     int mid=(first+end)>>1;
    35     build(lson);
    36     build(rson);
    37     push_up(rt);
    38     return ;
    39 }
    40 LL query(int rt,int first,int end,int a,int b) //求区间和
    41 {
    42        if(first>=a&&end<=b)
    43         return tree[rt];
    44     push_down(rt,end-first+1);
    45     LL sum=0;
    46     int mid=(first+end)>>1;
    47     if(a<=mid)
    48         sum+=query(lson,a,b);
    49     if(b>mid)
    50         sum+=query(rson,a,b);
    51     return sum;
    52 }
    53 void update(int rt,int first,int end,int a,int b,int c)
    54 {
    55     if(first>=a&&end<=b)
    56     {
    57         add[rt]+=c;
    58         tree[rt]+=(LL)c*(end-first+1);
    59         return ;
    60     }
    61     push_down(rt,end-first+1);
    62     int mid=(first+end)>>1;
    63     if(a<=mid)
    64         update(lson,a,b,c);
    65     if(b>mid)
    66         update(rson,a,b,c);
    67     push_up(rt);
    68     return ;
    69 }
    70 int main()
    71 {
    72     int n,q;
    73     char ch[5];
    74     int a,b,c;
    75     scanf("%d%d",&n,&q);
    76     //cin>>n>>q;
    77     build(1,1,n);
    78     while(q--)
    79     {
    80         scanf("%s",ch);
    81         //cin>>ch;
    82         if(ch[0]=='Q')
    83         {
    84             scanf("%d%d",&a,&b);
    85             printf("%lld
    ",query(1,1,n,a,b));
    86             //cin>>a>>b;
    87             //cout<<query(1,1,n,a,b)<<endl;
    88         }
    89         if(ch[0]=='C')
    90         {
    91             scanf("%d%d%d",&a,&b,&c);
    92             //cin>>a>>b>>c;
    93             update(1,1,n,a,b,c);
    94         }
    95     }
    96     return 0;
    97 }
    View Code
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  • 原文地址:https://www.cnblogs.com/cxbky/p/4923309.html
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