分块---A Simple Problem with Integers
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
分块：
#include
#include
#include
#define ll long long
#define N 100010
using namespace std;
ll a[N],sum[N],add[N];
int L[N],R[N],d;
int pos[N];
int n,m,t,l,r;
char op[3];
void change(int l,int r,long long d)
{
int p=pos[l],q=pos[r];
if(p==q){
for(int i=l;i<=r;i++) a[i]+=d;
sum[p]+=d*(r-l+1);
}//全都在一块中
else
{
for(int i=p+1;i<=q-1;i++) add[i]+=d;//维护中间段落
for(int i=l;i<=R[p];i++) a[i]+=d;
sum[p]+=d*(R[p]-l+1);
for(int i=L[q];i<=r;i++) a[i]+=d;
sum[q]+=d*(r-L[q]+1);
//暴力局部修改
}
}
ll ask(int l,int r)
{
int p=pos[l],q=pos[r];
ll ans=0;
if(p==q){
for(int i=l;i<=r;i++) ans+=a[i];
ans+=add[p]*(r-l+1);
}//全部都在一块中
else
{
for(int i=p+1;i<=q-1;i++)
ans+=sum[i]+add[i]*(R[i]-L[i]+1);
//累加中间段落
for(int i=l;i<=R[p];i++) ans+=a[i];
ans+=add[p]*(R[p]-l+1);
for(int i=L[q];i<=r;i++) ans+=a[i];
ans+=add[q]*(r-L[q]+1);
//暴力局部累加
}
return ans;
}
int main()
{
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
scanf("%lld",&a[i]);
t=sqrt(n);
for(int i=1;i<=t;i++){
L[i]=(i-1)*t+1;
R[i]=i*t;
}//标记每块左右
if(R[t] 参考：https://blog.csdn.net/qq_39897867/article/details/82809676
线段树：
#include
#include
#include
#include
#define eps 0.00000001
using namespace std;
int num[400000];
struct node
{
int l,r;
long long nsum;
long long add;
}seg[400000];
void Build(int i,int l,int r)
{
seg[i].l=l;
seg[i].r=r;
seg[i].add=0;
if(l==r)
{
seg[i].nsum=num[l];
return ;
}
int mid=l+((r-l)>>1);
Build(i<<1,l,mid);
Build(i<<1|1,mid+1,r);
seg[i].nsum=seg[i<<1].nsum+seg[i<<1|1].nsum;
}
void Add(int i,int l,int r,long long c)
{
if(seg[i].l==l&&seg[i].r==r)
{
seg[i].add+=c;
return ;
}
seg[i].nsum+=c*(r-l+1);
int mid=(seg[i].l+seg[i].r)>>1;
if(r<=mid) Add(i<<1,l,r,c);
else if(l>mid) Add(i<<1|1,l,r,c);
else
{
Add(i<<1,l,mid,c);
Add(i<<1|1,mid+1,r,c);
}
}
long long Query(int i,int a,int b)//查询a-b的总和
{
if(seg[i].l==a&&seg[i].r==b)
{
return seg[i].nsum+(b-a+1)*seg[i].add;
}
seg[i].nsum+=(seg[i].r-seg[i].l+1)*seg[i].add;
int mid=(seg[i].l+seg[i].r)>>1;
Add(i<<1,seg[i].l,mid,seg[i].add);
Add(i<<1|1,mid+1,seg[i].r,seg[i].add);
seg[i].add=0;
if(b<=mid) return Query(i<<1,a,b);
else if(a>mid) return Query(i<<1|1,a,b);
else return Query(i<<1,a,mid)+Query(i<<1|1,mid+1,b);
}
int main()
{
int n,m,i;
int a,b,c;
char s[10];
while(~scanf("%d%d",&n,&m))
{
for(i=1;i<=n;i++)
{
scanf("%d",&num[i]);
}
Build(1,1,n);
for(i=0;i 参考：https://blog.csdn.net/bless924295/article/details/51569413
树状数组
#include
#include
#include
#define lobit(x) x&-x
using namespace std;
long long t[2][100010],x,a[100010],sum[100010];
int n,m,l,r;
void add(int k,int x,int num)//修改
{
while(x<=n)
{
t[k][x]+=num;
x+=lobit(x);
}
}
long long ask(int k,int x)//查询
{
long long sum=0;
while(x>0)
{
sum+=t[k][x];
x-=lobit(x);
}
return sum;
}
int main()
{
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
{
scanf("%lld",&a[i]);
sum[i]=sum[i-1]+a[i];
}
for(int i=1;i<=m;i++)
{
char c[2];
scanf("%s %d %d",c,&l,&r);
if(c[0]=='Q')
{
long long ans=sum[r]+(r+1)*ask(0,r)-ask(1,r);
ans-=sum[l-1]+l*ask(0,l-1)-ask(1,l-1);
//查询
printf("%lld\n",ans);
}
else
{
scanf("%d",&x);
add(0,l,x);
add(0,r+1,-x);
add(1,l,l*x);
add(1,r+1,-(r+1)*x);//修改
}
}
}
参考：https://blog.csdn.net/Mr_wuyongcong/article/details/81974566