| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 92632 | Accepted: 28818 | |
| 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
线段树的区间更新，区间求和裸题，感觉线段树比树状数组好理解……先照着别的的模版敲一遍再理解，线段树的写法很多人都存在差异，改成自己的比较好用
代码：
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<cstdio>
#include<string>
#include<deque>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
#define INF 0x3f3f3f3f
#define MM(x,y) memset(x,y,sizeof(x))
#define LC(x) (x<<1)
#define RC(x) (x<<1)|1
typedef pair<int,int> pii;
typedef long long LL;
const double PI=acos(-1.0);
const int N=100010;
struct info
{
LL l,r,mid;
LL sum;
LL add;
}T[N<<2];
LL arr[N];
void pushup(int k)
{
T[k].sum=T[LC(k)].sum+T[RC(k)].sum;
}
void pushdown(int k)
{
T[LC(k)].add+=T[k].add;
T[LC(k)].sum+=T[k].add*(T[LC(k)].r-T[LC(k)].l+1);
T[RC(k)].add+=T[k].add;
T[RC(k)].sum+=T[k].add*(T[RC(k)].r-T[RC(k)].l+1);
T[k].add=0;
}
void build(int k,LL l,LL r)
{
T[k].l=l;
T[k].r=r;
T[k].add=0;
T[k].mid=(T[k].l+T[k].r)>>1;
if(l==r)
T[k].sum=arr[l];
else
{
build(LC(k),l,T[k].mid);
build(RC(k),T[k].mid+1,r);
pushup(k);
}
}
void update(LL l,LL r,LL val,int k)
{
if(r<T[k].l||l>T[k].r)
return ;
if(l<=T[k].l&&r>=T[k].r)
{
T[k].add+=val;
T[k].sum+=val*(T[k].r-T[k].l+1);
}
else
{
if(T[k].add)
pushdown(k);
update(l,r,val,LC(k));
update(l,r,val,RC(k));
pushup(k);
}
}
LL query(int k,LL l,LL r)
{
if(l<=T[k].l&&r>=T[k].r)
return T[k].sum;
if(T[k].add)
pushdown(k);
if(r<=T[k].mid)
return query(LC(k),l,r);
else if(l>T[k].mid)
return query(RC(k),l,r);
else
return query(LC(k),l,T[k].mid)+query(RC(k),T[k].mid+1,r);
}
int main(void)
{
int tcase,n,i,m;
LL l,r,val;
char ops[3];
while (~scanf("%d%d",&n,&m))
{
MM(arr,0);
for (i=1; i<=n; i++)
scanf("%lld",&arr[i]);
build(1,1,n);
for (i=0; i<m; i++)
{
scanf("%s",ops);
if(ops[0]=='Q')
{
scanf("%lld%lld",&l,&r);
printf("%lld\n",query(1,l,r));
}
else
{
scanf("%lld%lld%lld",&l,&r,&val);
update(l,r,val,1);
}
}
}
return 0;
}