【POJ 3468】A Simple Problem with Integers
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 68616 | Accepted: 21153 | |
| 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
Source
树状数组维护带修改的区间和~
首先这道题用线段树打标记很好做，但是树状数组可以更快的实现~
树状数组本身是不支持区间修改的，那么我们怎么办呢？
s[i]表示最开始的前缀和。
数组de[i]，表示i-n的所有数都加上de[i]。
当把区间l-r都加上c的时候，de[l]+=c,de[r+1]-=c
我们看看怎么计算修改后的前缀和（前x个数）
原始和+修改和
=s[x]+del[1]*x+del[2]*(x-1)+...+del[x]*1
=s[x]+sigma(del[i]*(x-i+1))
=s[x]+sigma((1+x)*del[i]-i*del[i]) (1<=i<=x)
=s[x]+(1+x)*sigma(del[i])-sigma(i*del[i]) (1<=i<=x)
那么我们可以用两个树状数组：
d维护del[i]的前缀和；
d[i]维护del[i]*i的前缀和。
然后求l-r的区间和的时候，用前缀和相减就可以了~
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#define LL long long
#define M 100000+5
using namespace std;
int n,T;
LL d[M],di[M],s[M];
int lowbit(int x)
{
return x&(-x);
}
void Update1(int x,LL k)
{
for (int i=x;i<=n;i+=lowbit(i))
d[i]+=k;
}
void Update2(int x,LL k)
{
for (int i=x;i<=n;i+=lowbit(i))
di[i]+=k;
}
LL Getsum1(int x)
{
LL ans=0;
for (int i=x;i;i-=lowbit(i))
ans+=d[i];
return ans;
}
LL Getsum2(int x)
{
LL ans=0;
for (int i=x;i;i-=lowbit(i))
ans+=di[i];
return ans;
}
int main()
{
scanf("%d%d",&n,&T);
s[0]=0;
LL x;
for (int i=1;i<=n;i++)
scanf("%I64d",&x),s[i]=s[i-1]+x;
while (T--)
{
char ch[3];
int l,r;
LL c;
scanf("%s",ch);
scanf("%d%d",&l,&r);
if (ch[0]=='C')
{
scanf("%I64d",&c);
Update1(l,c);
Update1(r+1,-c);
Update2(l,c*(LL)l);
Update2(r+1,(-c)*(LL)(r+1));
}
else
{
LL ans=0;
ans=s[r]-s[l-1];
ans+=(Getsum1(r)*(LL)(r+1)-Getsum2(r));
ans-=(Getsum1(l-1)*(LL)l-Getsum2(l-1));
printf("%I64d\n",ans);
}
}
return 0;
}
感悟：
1.WA是因为getsum1/2搞反。。POJ要用I64d