poj 3468 A Simple Problem with Integers 解题报告

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 43907		Accepted: 12862
Case Time Limit: 2000MS

Description

You have N integers, A₁, A₂, ... , 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 A₁, A₂, ... , 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

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

这和上一题基本上是一致的，只是在细节上注意，在QUERY（）函数里面，也是要把延迟的标记，更新的，一定要注意！些外这题是要用INT64的！

#include <iostream>
#include <stdio.h>
using namespace std;
#define N 111111
__int64 l[N<<2],flag[N<<2];
void build(__int64 num,__int64 s,__int64 e)
{
      flag[num]=0;
      if(s==e)
      {
            scanf("%I64d",&l[num]);
            return ;
      }
      __int64 mid=(s+e)>>1;
      build(num<<1,s,mid);
      build(num<<1|1,mid+1,e);
      l[num]=l[num<<1]+l[num<<1|1];
}
void update(__int64 num ,__int64 s,__int64 e,__int64 a, __int64 b,__int64 c)
{
      if(a<=s&&b>=e)
      {
            flag[num]+=c;

            l[num]+=(e-s+1)*c;
            return ;
      }
      if(flag[num])
      {
            flag[num<<1]+=flag[num];
            flag[num<<1|1]+=flag[num];
            l[num<<1]+=(e-s+1-((e-s+1)>>1))*flag[num];
            l[num<<1|1]+=((e-s+1)>>1)*flag[num];
            flag[num]=0;
      }
      __int64 mid=(s+e)>>1;
      if(mid>=a)
      {
            update(num<<1,s,mid,a,b,c);
      }
      if(b>mid)
      {
            update(num<<1|1,mid+1,e,a,b,c);

      }
      l[num]=l[num<<1]+l[num<<1|1];
}
__int64 query(__int64 num,__int64 s,__int64 e ,__int64 a,__int64 b)
{
      if(a<=s&&b>=e)
      {
            return l[num];
      }
      if(flag[num])
      {
            flag[num<<1]+=flag[num];
            flag[num<<1|1]+=flag[num];
            l[num<<1]+=(e-s+1-((e-s+1)>>1))*flag[num];
            l[num<<1|1]+=((e-s+1)>>1)*flag[num];
            flag[num]=0;
      }
      __int64 mid=(s+e)>>1;
      __int64 re=0;
      if(mid>=a)re+=query(num<<1,s,mid,a,b);
      if(mid<b) re+=query(num<<1|1,mid+1,e,a,b);
      return re;
}
int  main()
{
      __int64 n,q,a,b,d;
      char c;
    while(scanf("%I64d%I64d",&n,&q)!=EOF)
    {
          build(1,1,n);
          while(q--)
          {
                getchar();
                if((c=getchar())=='Q')
                {
                    scanf("%I64d%I64d",&a,&b);
                    printf("%I64d\n",query(1,1,n,a,b));

                }
                else if(c=='C')
                {
                      scanf("%I64d%I64d%I64d",&a,&b,&d);

                      update(1,1,n,a,b,d);
                }


          }
    }
    return 0;
}