标签：poj3468
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 57666 | Accepted: 17546 | |
| 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
//#define DEBUG
#include <stdio.h>
#define maxn 100002
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
long long tree[maxn << 2], arr[maxn], lazy[maxn << 2];
void pushUp(int rt)
{
tree[rt] = tree[rt << 1] + tree[rt << 1 | 1];
}
void pushDown(int l, int r, int rt)
{
int mid = (l + r) >> 1;
tree[rt << 1] += (mid - l + 1) * lazy[rt];
tree[rt << 1 | 1] += (r - mid) * lazy[rt];
lazy[rt << 1] += lazy[rt];
lazy[rt << 1 | 1] += lazy[rt];
lazy[rt] = 0;
}
void build(int l, int r, int rt)
{
lazy[rt] = 0;
if(l == r){
tree[rt] = arr[r]; return;
}
int mid = (l + r) >> 1;
build(lson);
build(rson);
pushUp(rt);
}
void update(int left, int right, long long val, int l, int r, int rt)
{
if(left == l && right == r){
lazy[rt] += val; tree[rt] += val * (r - l + 1); return;
} //include l == r
if(lazy[rt]) pushDown(l, r, rt);
int mid = (l + r) >> 1;
if(right <= mid) update(left, right, val, lson);
else if(left > mid) update(left, right, val, rson);
else{
update(left, mid, val, lson);
update(mid + 1, right, val, rson);
}
pushUp(rt);
}
long long query(int left, int right, int l, int r, int rt)
{
if(left == l && right == r)
return tree[rt];
if(lazy[rt]) pushDown(l, r, rt);
int mid = (l + r) >> 1;
if(right <= mid){
return query(left, right, lson);
}else if(left > mid){
return query(left, right, rson);
}else{
return query(left, mid, lson) + query(mid + 1, right, rson);
}
}
int main()
{
#ifdef DEBUG
freopen("../stdin.txt", "r", stdin);
freopen("../stdout.txt", "w", stdout);
#endif
int n, q, i, a, b;
long long c;
char com[2];
while(scanf("%d%d", &n, &q) == 2){
for(i = 1; i <= n; ++i)
scanf("%lld", arr + i);
build(1, n, 1);
while(q--){
scanf("%s%d%d", com, &a, &b);
if(com[0] == 'C'){
scanf("%lld", &c);
update(a, b, c, 1, n, 1);
}else printf("%lld\n", query(a, b, 1, n, 1));
}
}
return 0;
}POJ3468 A Simple Problem with Integers 【线段树】+【成段更新】,布布扣,bubuko.com
POJ3468 A Simple Problem with Integers 【线段树】+【成段更新】
标签：poj3468
原文地址：http://blog.csdn.net/chang_mu/article/details/37691631
