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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 abc" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q ab" 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
/*
线段树的区间更新
添加懒惰标记
*/
#include <cstdio>
#include <algorithm>
using namespace std;
const int MAX = 1000000;
long long a[MAX];
long long lazy[MAX];
long long sum[MAX<<2];
void build(long long rt, long long l, long long r)
{
sum[rt] = lazy[rt] = 0;
if(l == r){
sum[rt] = a[l];
return ;
}
long long mid = (l + r) /2 ;
build(rt*2, l, mid);
build(rt*2+1, mid+1, r);
sum[rt] = sum[rt*2] + sum[rt*2+1];
}
void down(long long rt, long long l, long long r)
{
if(lazy[rt]){
long long mid = (l + r) / 2;
lazy[rt*2] += lazy[rt]; lazy[rt*2+1] += lazy[rt];
sum[rt*2] += lazy[rt]*(mid-l+1);
sum[rt*2+1] += lazy[rt]*(r-mid);
lazy[rt] = 0;
}
}
void update(long long rt, long long l, long long r, long long L, long long R, long long y)
{
if(L <= l && R >= r){
lazy[rt] += y;
sum[rt] += y*(r-l+1);
return;
}
down(rt, l, r);
long long mid = (l + r) /2 ;
if(L <= mid) update(rt*2, l , mid, L, R, y);
if(R > mid) update(rt*2+1, mid+1, r, L, R, y);
sum[rt] = sum[rt*2] + sum[rt*2+1];
}
long long query(long long rt, long long l, long long r, long long L, long long R)
{
if(L <= l && R >= r) return sum[rt];
down(rt, l, r);
long long mid = (l + r) / 2;
long long ret = 0;
if(L <= mid) ret += query(rt*2, l , mid, L ,R);
if(R > mid) ret += query(rt*2+1, mid+1, r, L, R);
sum[rt] = sum[rt*2] + sum[rt*2+1];
return ret;
}
int main()
{
int n, q;
long long x, y,z;
char s[10];
while(~scanf("%d%d", &n, &q)){
for(int i = 1; i <= n ; i++)
scanf("%I64d", &a[i]);
build(1, 1, n);
for(int i = 1; i <= q; i++){
scanf("%s", s);
if(s[0] == ‘Q‘){
scanf("%I64d%I64d", &x, &y);
printf("%I64d\n", query(1, 1, n, x, y));
}
else{
scanf("%I64d%I64d%I64d", &x, &y, &z);
update(1, 1, n, x, y, z);
}
}
}
return 0;
}
POJ3468——线段树区间更新——A Simple Problem with Integers
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原文地址：http://www.cnblogs.com/zero-begin/p/4694198.html
