POJ - 3468 A Simple Problem with Integers 线段树
A Simple Problem with Integers
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 131119 | Accepted: 40685 | |
| 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
The sums may exceed the range of 32-bit integers.
Source
POJ Monthly--2007.11.25, Yang Yi
还是线段树的一道模板题
这题用了cin cout超时了QAQ，不知道是不是我代码的写的太丑
#include
#include
#include
using namespace std;
#define MAXN 200010
typedef long long ll;
ll n, m;
ll a[MAXN];
struct node{
ll l, r, tag,sum;
}tree[MAXN << 2];
char ch[2];
void build(ll k, ll l, ll r)
{
tree[k].l = l; tree[k].r = r;
if (l == r)
{
tree[k].sum = a[l];
return;
}
ll mid = (l + r) >> 1;
build(k << 1, l, mid); build(k << 1 | 1, mid + 1, r);
tree[k].sum = tree[k << 1].sum+ tree[k << 1 | 1].sum;
}
void prework()
{
for (ll i = 1; i <= n; i++)
scanf("%lld", &a[i]);
build(1, 1, n);
}
void change(ll k)
{
if (tree[k].l == tree[k].r)
{
tree[k].sum += tree[k].tag;
}
else
{
tree[k].sum += (tree[k].r - tree[k].l + 1)*tree[k].tag;
tree[k << 1].tag += tree[k].tag;
tree[k << 1 | 1].tag += tree[k].tag;
}
tree[k].tag = 0;
}
void add(ll k, ll l, ll r, ll x)
{
if (tree[k].tag)
change(k);
if (tree[k].l == l&&tree[k].r == r)
{
tree[k].tag += x;
return;
}
tree[k].sum += (r - l + 1)*x;
ll mid = (tree[k].l + tree[k].r) >> 1;
if (r <= mid)
add(k << 1, l, r, x);
else if (l >= mid+1)
add(k << 1 | 1, l, r, x);
else
{
add(k << 1, l, mid, x);
add(k << 1|1, mid + 1, r, x);
}
}
ll query(ll k, ll l, ll r)
{
if (tree[k].tag)
change(k);
ll mid = (tree[k].l + tree[k].r) >> 1;
if (tree[k].r == r&&tree[k].l == l)
return tree[k].sum;
if (l >= mid + 1)
return query(k << 1 | 1, l, r);
else if (r <= mid)
return query(k << 1, l, r);
else
return query(k << 1, l, mid) + query(k << 1|1, mid+1, r);
}
void mainwork()
{
ll num,d, x;
for (ll i = 1; i <= m; i++)
{
scanf("%s", ch);
if (ch[0] == 'Q')
{
scanf("%lld%lld", &d, &x);
printf("%lld\n", query(1, d, x));
}
else
{
scanf("%lld%lld%lld",&d,&x,&num);
add(1, d, x, num);
}
}
}
int main()
{
while (~scanf("%lld%lld", &n, &m)){
prework();
mainwork();
}
return 0;
}