Ryuji doesn’t want to study
问题分析
我们可以做一个前缀和sum{a[i]}
s
u
m
{
a
[
i
]
}
和一个前缀和sum{(n−i)∗a[i]}
s
u
m
{
(
n
−
i
)
∗
a
[
i
]
}
(看起来就像一个三角形，我们这里称之为tangle[i]
t
a
n
g
l
e
[
i
]
)，如图所示。
假设我们要求区间[2,5]=2⋅4+3⋅3+4⋅2+5⋅1
[
2
,
5
]
=
2
⋅
4
+
3
⋅
3
+
4
⋅
2
+
5
⋅
1
，那么，我们通过图中可以发现答案就是梯形的数值减去平行四边形的数值。即
query(2,5)=(tangle[5]−tangle[1])−(7−5)⋅(sum[5]−sum[1]) q u e r y ( 2 , 5 ) = ( t a n g l e [ 5 ] − t a n g l e [ 1 ] ) − ( 7 − 5 ) ⋅ ( s u m [ 5 ] − s u m [ 1 ] )
因此我们只需要用线段树维护两个前缀和即可。
这里蒻再补充一个做法。就是线段树维护前缀和的区间和。
我们将题里的要求分解一下。
假设仍然要求query(2,5)
q
u
e
r
y
(
2
,
5
)
，
分解开来就会发现
query(2,5)=(2+3+4+5)+(2+3+4)+(2+3)+(2)=(sum[5]−sum[1])+(sum[4]−sum[1])+(sum[3]−sum[1])+(sum[2]−sum[1])=∑i=25sum(i)−(5−3+1)⋅sum(1)(1)(2)(3) (1) q u e r y ( 2 , 5 ) = ( 2 + 3 + 4 + 5 ) + ( 2 + 3 + 4 ) + ( 2 + 3 ) + ( 2 ) (2) = ( s u m [ 5 ] − s u m [ 1 ] ) + ( s u m [ 4 ] − s u m [ 1 ] ) + ( s u m [ 3 ] − s u m [ 1 ] ) + ( s u m [ 2 ] − s u m [ 1 ] ) (3) = ∑ i = 2 5 s u m ( i ) − ( 5 − 3 + 1 ) ⋅ s u m ( 1 )
所以题目所求变为
query(l,r)=∑i=lrsum(i)−(r−l+1)⋅sum(l−1) q u e r y ( l , r ) = ∑ i = l r s u m ( i ) − ( r − l + 1 ) ⋅ s u m ( l − 1 )
代码(第一种做法)
#include <cstdio>
using namespace std;
typedef long long LL;
const int N = 1e6+4;
int n,q,a[N];
struct data {
int l, r;
LL val, tangle;
} tr[N<<1];
inline void pushup(int rt) {
tr[rt].val = tr[rt << 1].val + tr[rt << 1 | 1].val;
tr[rt].tangle = tr[rt << 1].tangle + tr[rt << 1 | 1].tangle;
}
void build(int k,int s,int t) {//建树
tr[k].l = s;
tr[k].r = t;
if (s == t) {
tr[k].val = a[s];
tr[k].tangle = a[s] * 1LL * (n - s + 1);
return;
}
int mid = (s + t) >> 1;
build(k << 1, s, mid);
build(k << 1 | 1, mid + 1, t);
pushup(k);
}
LL query(int k,int x,int y,int len) {//区间求和
int L = tr[k].l, R = tr[k].r;
if (x == L && y == R)
return tr[k].tangle - 1LL * len * tr[k].val;
int mid = (L + R) >> 1;
if (y <= mid)
return query(k << 1, x, y, len);
else if (x > mid)
return query(k << 1 | 1, x, y, len);
else
return query(k << 1, x, mid, len) + query(k << 1 | 1, mid + 1, y, len);
}
void update(int rt,int L,int R,int x,LL v) {
if (L == R) {
tr[rt].tangle = (n - x + 1) * 1LL * v;
tr[rt].val = v;
return;
}
int mid = (L + R) >> 1;
if (x <= mid)
update(rt << 1, L, mid, x, v);
else
update(rt << 1 | 1, mid + 1, R, x, v);
pushup(rt);
}
int main() {
scanf("%d%d", &n, &q);
for (int i = 1; i <= n; i++)
scanf("%d", &a[i]);
build(1, 1, n);
int op, x, y;
for (int i = 1; i <= q; i++) {
scanf("%d%d%d", &op, &x, &y);
if (op == 2)
update(1, 1, n, x, y);
else
printf("%lld\n", query(1, x, y, n - y));
}
return 0;
}
代码(第二种做法)
#include <cstdio>
#define L tree[rt].l
#define R tree[rt].r
typedef long long LL;
using namespace std;
const int N = 1e6+4;
int a[N],n,q;
LL sum[N];
struct data {
int l,r;
LL Inc,nSum;
} tree[N];
void pushup(int rt)
{
tree[rt].nSum = tree[rt<<1].nSum+tree[rt<<1|1].nSum;
}
void pushdown(int rt) {
if(tree[rt].Inc) {
int mid = (L+R)>>1;
tree[rt<<1].Inc += tree[rt].Inc;
tree[rt<<1|1].Inc += tree[rt].Inc;
tree[rt<<1].nSum += tree[rt].Inc*(mid-L+1);
tree[rt<<1|1].nSum += tree[rt].Inc*(R-mid);
tree[rt].Inc = 0;
}
}
void build(int rt,int x,int y) {//建树
tree[rt].l = x; tree[rt].r = y;
if(x == y) {
tree[rt].Inc = 0;
tree[rt].nSum = sum[x];
return;
}
int mid = (x + y) >> 1;
build(rt<<1,x,mid);
build(rt<<1|1,mid+1,y);
pushup(rt);
}
LL query(int rt,int x,int y){//区间求和
if(x == L && y == R)
return tree[rt].nSum;
pushdown(rt);
int mid=(L+R)>>1;
if(y<=mid)
return query(rt<<1,x,y);
else if(x>mid)
return query(rt<<1|1,x,y);
else
return query(rt<<1,x,mid)+query(rt<<1|1,mid+1,y);
}
void update(int rt,int x,int y,LL v)
{
if(x == L && R == y) {
tree[rt].nSum += v*1LL*(R-L+1);
tree[rt].Inc += v;
return;
}
pushdown(rt);
int mid = (L+R)>>1;
if(y <= mid)
update(rt << 1, x, y,v);
else if(x > mid)
update(rt << 1 | 1, x, y,v);
else{
update(rt << 1, x, mid,v);
update(rt << 1 | 1, mid+1, y,v);
}
pushup(rt);
}
int main(){
scanf("%d%d",&n,&q);
for(int i=1; i<=n; i++) {
scanf("%d",&a[i]);
sum[i] = sum[i-1] + a[i];
}
build(1,1,n);
int op,x,y;
for(int i=1; i<=q; i++) {
scanf("%d%d%d",&op,&x,&y);
if(op==2) {
update(1,x,n,1LL*(y-a[x]));
a[x] = y;
}
else{
if(x==1) {
printf("%lld\n", query(1,x,y));
}else
printf("%lld\n", query(1,x,y)-1LL*(y-x+1)*query(1,x-1,x-1));
}
}
return 0;
}
