poj 3468 A Simple Problem with Integers 线段树区间加，区间查询和

A Simple Problem with Integers

Time Limit: 1 Sec  Memory Limit: 256 MB

题目连接

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

Description

You have N integers, A ₁, A ₂, ... , 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 A ₁, A ₂, ... , 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

 

4
55
9
15

 

HINT

 

 

 

题意

 

 区间加，区间查询和

 

题解：

 

线段树，记住懒操作~

 

代码：

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 200001
#define mod 10007
#define eps 1e-9
int Num;
char CH[20];
//const int inf=0x7fffffff;   //нчоч╢С
const int inf=0x3f3f3f3f;
/*

inline void P(int x)
{
    Num=0;if(!x){putchar('0');puts("");return;}
    while(x>0)CH[++Num]=x%10,x/=10;
    while(Num)putchar(CH[Num--]+48);
    puts("");
}
*/
//**************************************************************************************
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
inline void P(int x)
{
    Num=0;if(!x){putchar('0');puts("");return;}
    while(x>0)CH[++Num]=x%10,x/=10;
    while(Num)putchar(CH[Num--]+48);
    puts("");
}
struct node
{
    int l,r;
    ll sum,add;
    void fun(ll tmp)
    {
        add+=tmp;
        sum+=(r-l+1)*tmp;
    }
}a[maxn*4];
ll d[maxn];
void relax(int x)
{
    if(a[x].add)
    {
        a[x<<1].fun(a[x].add);
        a[x<<1|1].fun(a[x].add);
        a[x].add=0;
    }
}
void build(int x,int l,int r)
{
    a[x].l=l,a[x].r=r;
    if(l==r)
    {
        a[x].sum=d[l];
    }
    else
    {
        int mid=(l+r)>>1;
        build(x<<1,l,mid);
        build(x<<1|1,mid+1,r);
        a[x].sum=a[x<<1].sum+a[x<<1|1].sum;
    }
}
void update(int x,int st,int ed,ll c)
{
    int l=a[x].l,r=a[x].r;
    if(st<=l&&r<=ed)
        a[x].fun(c);
    else
    {
        relax(x);
        int mid=(l+r)>>1;
        if(st<=mid)update(x<<1,st,ed,c);
        if(ed>mid) update(x<<1|1,st,ed,c);
        a[x].sum=a[x<<1].sum+a[x<<1|1].sum;
    }
}
ll query(int x,int st,int ed)
{
    int l=a[x].l,r=a[x].r;
    if(st<=l&&r<=ed)
        return a[x].sum;
    else
    {
        relax(x);
        int mid=(l+r)>>1;
        ll sum1=0,sum2=0;
        if(st<=mid)
            sum1=query(x<<1,st,ed);
        if(ed>mid)
            sum2=query(x<<1|1,st,ed);
        return sum1+sum2;
    }
}
int main()
{
    int n=read(),m=read();
    for(int i=1;i<=n;i++)
        d[i]=read();
    build(1,1,n);
    char s[3];
    int bb,cc,dd;
    for(int i=0;i<m;i++)
    {
        scanf("%s",s);
        if(s[0]=='Q')
        {
            bb=read(),cc=read();
            printf("%lld\n",query(1,bb,cc));
        }
        else
        {
            bb=read(),cc=read(),dd=read();
            update(1,bb,cc,dd);
        }
    }
}

 

D. Points

Time Limit: 1 Sec  Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/19/problem/D

Description

Pete and Bob invented a new interesting game. Bob takes a sheet of paper and locates a Cartesian coordinate system on it as follows: point (0, 0) is located in the bottom-left corner, Ox axis is directed right, Oy axis is directed up. Pete gives Bob requests of three types:

• add x y — on the sheet of paper Bob marks a point with coordinates (x, y). For each request of this type it's guaranteed that point (x, y) is not yet marked on Bob's sheet at the time of the request.
• remove x y — on the sheet of paper Bob erases the previously marked point with coordinates (x, y). For each request of this type it's guaranteed that point (x, y) is already marked on Bob's sheet at the time of the request.
• find x y — on the sheet of paper Bob finds all the marked points, lying strictly above and strictly to the right of point (x, y). Among these points Bob chooses the leftmost one, if it is not unique, he chooses the bottommost one, and gives its coordinates to Pete.

Bob managed to answer the requests, when they were 10, 100 or 1000, but when their amount grew up to 2·10⁵, Bob failed to cope. Now he needs a program that will answer all Pete's requests. Help Bob, please!

Input

The first input line contains number n (1 ≤ n ≤ 2·10⁵) — amount of requests. Then there follow n lines — descriptions of the requests. add x y describes the request to add a point, remove x y — the request to erase a point, find x y — the request to find the bottom-left point. All the coordinates in the input file are non-negative and don't exceed 10⁹.

Output

For each request of type find x y output in a separate line the answer to it — coordinates of the bottommost among the leftmost marked points, lying strictly above and to the right of point (x, y). If there are no points strictly above and to the right of point (x, y), output -1.

 

Sample Input

7

add 1 1

add 3 4

find 0 0

remove 1 1

find 0 0

add 1 1

find 0 0

Sample Output

1 1

3 4

1 1

HINT

 

题意

 二维插入删除，找右上角的点

题解：

线段树套set

代码：