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
线段树区间更新。
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
#define lson o << 1, l, m
#define rson o << 1|1, m+1, r
using namespace std;
typedef long long LL;
const int MAX=0x3f3f3f3f;
const int maxn = 111111;
int n, q, a, b, c;
LL sum[maxn<<2], add[maxn<<2];
void up(int o) {
sum[o] = sum[o<<1] + sum[o<<1|1];
}
void down(int o, int m) {
if(add[o] != 0) {
add[o<<1] += add[o];
add[o<<1|1] += add[o];
sum[o<<1] += add[o]*(m-(m>>1));
sum[o<<1|1] += add[o]*(m>>1);
add[o] = 0;
}
}
void build(int o, int l, int r) {
add[o] = 0;
if(l == r) scanf("%I64d", &sum[o]);
else {
int m = (l+r) >> 1;
build(lson);
build(rson);
up(o);
}
}
LL query(int o, int l, int r) {
if(a <= l && r <= b) return sum[o];
down(o, r-l+1);
int m = (l+r) >> 1;
LL ans = 0;
if(a <= m) ans += query(lson);
if(m < b) ans += query(rson);
return ans;
}
void update(int o, int l, int r) {
if(a <= l && r <= b) {
add[o] += c;
sum[o] += (LL)c*(r-l+1);
return;
}
int m = (l+r) >> 1;
down(o, r-l+1);
if(a <= m) update(lson);
if(m < b) update(rson);
up(o);
}
int main()
{
scanf("%d%d", &n, &q);
build(1, 1, n);
while(q--) {
char op[3];
scanf("%s",op);
if(op[0] == 'C') {
scanf("%d%d%d", &a, &b, &c);
update(1, 1, n);
} else {
scanf("%d%d", &a, &b);
printf("%I64d\n", query(1, 1, n));
}
}
return 0;
}
POJ 3468 A Simple Problem with Integers（线段树 区间更新）,布布扣,bubuko.com
POJ 3468 A Simple Problem with Integers（线段树 区间更新）
原文地址：http://blog.csdn.net/u013923947/article/details/38380765
