POJ 3468 区间更新（求任意区间和）A Simple Problem with Integers
A Simple Problem with Integers
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 163977 | Accepted: 50540 | |
| 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
#include<iostream>
#include<algorithm>
#include<string.h>
#include<string>
#define ll long long
using namespace std;
ll tree[], lazy[], len[];//tree[num]存的是節點num所在區間的區間和
void pushdown(ll num)
{
if (lazy[num] != )
{
tree[num * ] = tree[num * ] + lazy[num] * len[num * ];
tree[num * + ] = tree[num * + ] + lazy[num] * len[num * + ];
lazy[num * ] = lazy[num * ] + lazy[num];
lazy[num * + ] = lazy[num * + ] + lazy[num];
lazy[num] = ;
}
} void build(ll num, ll le, ll ri)
{
len[num] = ri - le + ;//區間長度
if (le == ri)
{
scanf("%lld", &tree[num]);
return;
}
ll mid = (le + ri) / ;
build(num * , le, mid);
build(num * + , mid + , ri);
tree[num] = tree[num * ] + tree[num * + ];
} void update(ll num, ll le, ll ri, ll x, ll y, ll z)
{
if (x <= le && ri <= y)
{
lazy[num] = lazy[num] + z;
tree[num] = tree[num] + len[num] * z;//更新區間和
return;
}
pushdown(num);
ll mid = (le + ri) / ;
if (x <= mid)
update(num * , le, mid, x, y, z);
if (y > mid)
update(num * + , mid + , ri, x, y, z);
tree[num]=tree[num*]+tree[num*+];
} ll query(ll num, ll le, ll ri, ll x, ll y)
{
if (x <= le && ri <= y)//查詢區間在num節點所在區間內
return tree[num];
pushdown(num);
ll mid = (le + ri) / ;
ll ans = ;
if (x <= mid)
ans = ans + query(num * , le, mid, x, y);
if (y > mid)
ans = ans + query(num * + , mid + , ri, x, y);
return ans;
}
int main()
{
ll n, m;
scanf("%lld%lld", &n, &m);
build(, , n);
while (m--)
{
char c[];
scanf("%s", c);
if (c[] == 'Q')
{
ll x, y;
scanf("%lld%lld", &x, &y);
printf("%lld\n", query(, , n, x, y));
}
else
{
ll x, y, z;
scanf("%lld%lld%lld", &x, &y, &z);
update(, , n, x, y, z);
}
}
return ;
}
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