首页 > 代码库 > poj 3468 A Simple Problem with Integers

poj 3468 A Simple Problem with Integers

2024-07-16 03:36:18   232人阅读

A Simple Problem with Integers
Time Limit: 5000MS Memory Limit: 131072K
Total Submissions: 60604 Accepted: 18470
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 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




线段树（区间求和，更新线段）

#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std; 
#define MAX_N 100001  

__int64 sum = 0;  
__int64 num[MAX_N];  

struct node  
{  
    int left;  
    int right;  
    __int64 count; //表示该节点的总个数  
    __int64 data;  //表示增量的那个数  
}N[4*MAX_N];  


void CreateTree(int i,int a,int b)  
{  
    N[i].left = a;  
    N[i].right = b;  
    N[i].data = http://www.mamicode.com/0;  >
TML代码：
真心苦逼啊。。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

int max(int a,int b)
{
    return (a>b?a:b);
}
int a[200000];
struct Node
{
    int left,right;
    int t;
}node[4*200000];

void MakeTree(int l,int r,int i)
{
  node[i].left=l;
  node[i].right=r;

   if(l==r)
   {
       node[i].t=a[l];
       return ;
   }
  int mid=(l+r)/2;
  MakeTree(l,mid,2*i);
  MakeTree(mid+1,r,2*i+1);
  node[i].t=node[2*i].t+node[2*i+1].t;
}

void UpdateTree(int i,int x,int k)
{
    int l=node[i].left;
    int r=node[i].right;
    int mid=(l+r)/2;
    if(x==l&&x==r)
    {
        node[i].t+=k;
        return ;
    }
    if(x>mid)
        UpdateTree(2*i+1,x,k);
    else
        UpdateTree(2*i,x,k);

    node[i].t=node[2*i].t+node[2*i+1].t;

}

int QueryTree(int x,int y,int i)
{
    if(node[i].left==x && node[i].right==y)
        return node[i].t;
    int m=(node[i].left+node[i].right)/2;

    if(x>m)
        return QueryTree(x,y,2*i+1);
    else if(y<=m)
        return QueryTree(x,y,2*i);
    else
        return QueryTree(x,m,2*i)+QueryTree(m+1,y,2*i+1);
}

int main ()
{
    int T,n,m;
    int i,j,x,y,k;
    char c[10];
    while(~scanf("%d%d",&n,&m))
    {

        for(i=1;i<=n;i++)
            scanf("%d",&a[i]);

        MakeTree(1,n,1);
        for(j=1;j<=m;j++)
        {
         scanf("%s%d%d",c,&x,&y);
         if(c[0]=='C')
         {
             scanf("%d",&k);
             for(i=x;i<=y;i++)
             {
                 UpdateTree(1,i,k);
             }
         }
         else
            cout<<QueryTree(x,y,1)<<endl;
       }
    }
    return 0;
}






联系
我们