【nowcoder】Tree Recovery（线段树-区间更新 / 前缀和与差分）

链接：https://www.nowcoder.com/acm/contest/77/H

来源：牛客网

时间限制：C/C++ 1秒，其他语言2秒

空间限制：C/C++ 131072K，其他语言262144K

64bit IO Format: %lld

题目描述

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.

输入描述:

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.

输出描述:

You need to answer all Q commands in order. One answer in a line.

示例1

输入

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

输出

4

55

9

15

分析：题目数据比较弱，可以直接暴力求解

另外可用一维前缀和与差分 复杂度 O（n）

当数据较大，可用线段树的区间更新求和 模板

【前缀和与差分】代码：

/**

给定序列a，对[l,r] 的所有元素都加上k，求[L,R]的ai和

用前缀和求解 O(n)

///p[i]=a[i]-a[i-1]

对【l，r】内元素加k 转化为 p[l]+=k,p[r+1]-=k;

然后对修改后的数组a求和

一旦n过大，还是不适用

**/

#include <bits/stdc++.h>
using namespace std;

#define mem(a,n) memset(a,n,sizeof(a))
#define memc(a,b) memcpy(a,b,sizeof(b))
#define rep(i,a,n) for(int i=a;i<n;i++) ///[a,n)
#define dec(i,n,a) for(int i=n;i>=a;i--)///[n,a]
#define pb push_back
#define fi first
#define se second
#define IO ios::sync_with_stdio(false)
#define fre freopen("in.txt","r",stdin)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
typedef long long ll;
typedef unsigned long long ull;
const double PI=acos(-1.0);
const double E=2.718281828459045;
const double eps=1e-3;
const int INF=0x3f3f3f3f;
const int MOD=258280327;
const int N=1e5+5;
const ll maxn=1e6+5;
const int dir[4][2]= {-1,0,1,0,0,-1,0,1};
int a
,p
;
int n,q;
void add(int l,int r,int k)
{
p[l]+=k,p[r+1]-=k;
}
void get_a()
{
rep(i,1,n+1) a[i]=a[i-1]+p[i];
}
int main()
{
while(~scanf("%d%d",&n,&q))
{
mem(a,0);
mem(p,0);
rep(i,1,n+1)
{
scanf("%d",&a[i]);
p[i]=a[i]-a[i-1];
}
while(q--)
{
char op;
scanf(" %c",&op);
if(op=='C')
{
int l,r,k;
scanf("%d%d%d",&l,&r,&k);
add(l,r,k);
get_a();
}
else
{
int l,r;
scanf("%d%d",&l,&r);
ll sum=0;
rep(i,l,r+1) sum+=a[i];
printf("%lld\n",sum);
}
}
}
return 0;
}

【线段树 区间更新模板】

#include <bits/stdc++.h>
using namespace std;

#define mem(a,n) memset(a,n,sizeof(a))
#define memc(a,b) memcpy(a,b,sizeof(b))
#define rep(i,a,n) for(int i=a;i<n;i++) ///[a,n)
#define dec(i,n,a) for(int i=n;i>=a;i--)///[n,a]
#define pb push_back
#define fi first
#define se second
#define IO ios::sync_with_stdio(false)
#define fre freopen("in.txt","r",stdin)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
typedef long long ll;
typedef unsigned long long ull;
const double PI=acos(-1.0);
const double E=2.718281828459045;
const double eps=1e-3;
const int INF=0x3f3f3f3f;
const int MOD=258280327;
const int N=1e5+5;
const ll maxn=1e6+5;
const int dir[4][2]= {-1,0,1,0,0,-1,0,1};
ll sum[N<<2],a
,MAX[N<<2],ad[N<<2];
void pushup(ll rt)
{
sum[rt]=sum[rt<<1]+sum[rt<<1|1];
// MAX[rt]=max(MAX[rt<<1],MAX[rt<<1|1]);
}
void pushdown(int rt,int m)
{
if(ad[rt])
{
ad[rt<<1]+=ad[rt];
ad[rt<<1|1]+=ad[rt];
sum[rt<<1]+=ad[rt]*(m-(m>>1));
sum[rt<<1|1]+=ad[rt]*(m>>1);
ad[rt]=0; ///标记清除！！！
}
}
void build(ll l,ll r,ll rt)
{
ad[rt]=0;
if(l==r)
{
scanf("%lld",&sum[rt]);
return ;
}
ll m=(l+r)>>1;
build(lson);
build(rson);
pushup(rt);
}
void update(ll L,ll R,ll c,ll l,ll r,ll rt)
{
if(L<=l&&R>=r)
{
ad[rt]+=c;
sum[rt]+=c*(r-l+1);
return ;
}
pushdown(rt,r-l+1);
ll m=(l+r)>>1;
if(L<=m) update(L,R,c,lson);
if(R>m) update(L,R,c,rson);
pushup(rt);
}
ll query(ll L,ll R,ll l,ll r,ll rt)
{
if(L<=l&&R>=r)
{
return sum[rt];
}
pushdown(rt,r-l+1);
ll m=(l+r)>>1;
ll ret=0;
if(L<=m) ret+=query(L,R,lson);
if(R>m) ret+=query(L,R,rson);
return ret;
}
int main()
{
int n,m;
while(~scanf("%d%d",&n,&m))
{
build(1,n,1);
for(int i=0; i<m; i++)
{
char op;
ll a,b,c;
scanf(" %c",&op);
if(op=='Q')
{
scanf("%lld%lld",&a,&b);
printf("%lld\n",query(a,b,1,n,1));

}
else
{
scanf("%lld%lld%lld",&a,&b,&c);
update(a,b,c,1,n,1);
}
}
}
return 0;
}