Description

You have N integers, A₁, A₂, ... , A_N. 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, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a+1, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a, A_a+1, ... , A_b.

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

其實最開始我調了很多次，並且這個程式過不了，因為我懶得改成int64了，留著自己以後看看

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
struct segtree{
long long l,r,v;
}tree[500005];
long long ans,add[500005];
void build(long long l,long long r,long long  root)
{
long long  m=(l+r)/2;
tree[root].l=l;
tree[root].r=r;
tree[root].v=0;
if(l==r)return;
build(l,m,2*root);
build(m+1,r,2*root+1);
}
void adjust(long long root,long long m)
{
if(add[root])
{
add[root*2]+=add[root];
add[root*2+1]+=add[root];
tree[root*2].v+=add[root]*(m-(m/2));
tree[root*2+1].v+=add[root]*(m/2);
add[root]=0; 
}
}
void insert(long long v,long long l,long long  root)
{
long long  m;
if(tree[root].l==tree[root].r&&tree[root].l==l)
{
tree[root].v+=v;
return;
}
m=(tree[root].l+tree[root].r)/2;
if(l<=m)insert(v,l,2*root);
else insert(v,l,2*root+1);
tree[root].v+=v;
}
void query(long long l,long long r,long long  root)
{
long long  m=(tree[root].l+tree[root].r)/2;
adjust(root,tree[root].r-tree[root].l+1);//每次查詢時，把lazy標籤處理了 
if(tree[root].l==l&&tree[root].r==r)
{
ans+=tree[root].v;
return;
}
if(m>=r)query(l,r,2*root);
else if(m<l)query(l,r,2*root+1);
else{
query(l,m,2*root);
query(m+1,r,2*root+1);
}
}
void update(long long c,long long l,long long r,long long root)
{
if(tree[root].l==l&&tree[root].r==r)
{
add[root]+=c;
tree[root].v+=c*(r-l+1);//區間和加這麼多 
return;
}
if(tree[root].l==tree[root].r)return;
adjust(root,tree[root].r-tree[root].l+1);
long long m=(tree[root].l+tree[root].r)/2;
if(r<=m)update(c,l,r,root*2);
else if(l>m)update(c,l,r,root*2+1);
else{
update(c,l,m,root*2);
update(c,m+1,r,root*2+1);
}
tree[root].v+=tree[root*2].v+tree[root*2+1].v;
}
int main()
{
long long  i,n,x,t,j;
cin>>n>>t;
build(1,n,1);
for(i=1;i<=n;i++)
{
cin>>j;
insert(j,i,1);;
}
char s;
while(t--)
{
cin>>s;
long long a,b,c;
if(s=='C')
{
cin>>a>>b>>c;
update(c,a,b,1);
}
else 
{
cin>>a>>b;
ans=0;
query(a,b,1);
cout<<ans<<endl;
}
}
return 0;
}

這是ac程式碼

#include<cstring>
#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
const int maxn=111111;
LL add[maxn<<2],sum[maxn<<2];
void pushup(int rt){
	sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
void build(int l,int r,int rt)
{
	add[rt]=0;
	if(l==r)
	{
		scanf("%lld",&sum[rt]);
		return;
	}
	int m=(l+r)>>1;
	build(lson);
	build(rson);
	pushup(rt);
}
void pushdown(int rt,int m)
{
	if(add[rt]){
		add[rt<<1]+=add[rt];
		add[rt<<1|1]+=add[rt];
		sum[rt<<1]+=add[rt]*(m-(m>>1));
		sum[rt<<1|1]+=add[rt]*(m>>1);
		add[rt]=0;
	}
}
void update(int x,int y,int c,int l,int r,int rt)
{
	if(x<=l&&y>=r)
	{
		add[rt]+=c;
		sum[rt]+=(LL)c*(r-l+1);
		return;
	}
	pushdown(rt,r-l+1);
	int m=(l+r)>>1;
	if(x<=m)update(x,y,c,lson);
	if(y>m)update(x,y,c,rson);
	pushup(rt);
}
LL query(int x,int y,int l,int r,int rt)
{
	if(x<=l&&y>=r)return sum[rt];
	pushdown(rt,r-l+1);
	int m=(l+r)>>1;
	LL ans=0;
	if(x<=m)ans+=query(x,y,lson);
	if(y>m)ans+=query(x,y,rson);
	return ans;
}
int main()
{
	int n,q;
	scanf("%d%d",&n,&q);
	build(1,n,1);
	while(q--)
	{
		char op[2];
		int a,b,c;
		scanf("%s",op);
		if(op[0]=='Q'){
			scanf("%d%d",&a,&b);
			printf("%lld\n",query(a,b,1,n,1));
		}
		else {
			scanf("%d%d%d",&a,&b,&c);
			update(a,b,c,1,n,1);
		}
	}
}