poj 3468 A Simple Problem with Integers 解题报告　线段树　数状数组两种实现

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, ... , 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

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;
}

这题用树状数组来做，更加简单的多，原理当然还是一样，只是树状数组写的更简单，毫无疑问，用树状数组，更加省空间，不会出现像线段树那样一搞就爆了内存！

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
#define MAXN 50005
int prime[MAXN],tree[12][12][MAXN],n;
int lowbit(int x)
{
return x&(-x);
}
int update(int i,int j,int x,int c)
{
if(x==0)
return -1;
while(x<=n)
{
tree[i][j][x]+=c;
x=x+lowbit(x);
}

return -1;
}
int getsum(int i,int j,int x)
{
int sum=0;
while(x>0)
{
sum+=tree[i][j][x];
x=x-lowbit(x);
}
return sum;
}
int main ()
{
int i,asknum,ask,pos,sum,a,b,k,c;
while(scanf("%d",&n)!=EOF)
{
memset(tree,0,sizeof(tree));
for(i=1;i<=n;i++)
{
scanf("%d",&prime[i]);
}
scanf("%d",&asknum);
while(asknum--)
{
scanf("%d",&ask);
if(ask==1)
{
scanf("%d%d%d%d",&a,&b,&k,&c);
int kk=(b-a)/k;
update(k,a%k,a,c);
update(k,a%k,b+1,-c);
}
else
{
scanf("%d",&pos);
sum=prime[pos];
for(i=1;i<=10;i++)
{
sum+=getsum(i,pos%i,pos);
}
printf("%d\n",sum);
}
}

}
return 0;
}