标签：des style os io for ar div line
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 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
Hint
题意：区间修改，区间求和
思路：典型的线段树区间求和
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define lson(x) ((x) << 1)
#define rson(x) ((x) << 1 | 1)
#define ll long long
using namespace std;
const int maxn = 100005;
struct seg {
ll w;
int flag;
};
struct segment_tree {
seg node[maxn<<2];
void update(int pos) {
node[pos].w = node[lson(pos)].w + node[rson(pos)].w;
}
void build(int l, int r, int pos) {
node[pos].flag = 0;
if (l == r) {
scanf("%lld", &node[pos].w);
return;
}
int m = l + r >> 1;
build(l, m, lson(pos));
build(m+1, r, rson(pos));
update(pos);
}
void push(int l, int r, int pos) {
int len = r - l + 1;
if (node[pos].flag != 0) {
node[lson(pos)].flag += node[pos].flag;
node[rson(pos)].flag += node[pos].flag;
node[lson(pos)].w += (ll)node[pos].flag * (len - (len >> 1));
node[rson(pos)].w += (ll)node[pos].flag * (len >> 1);
node[pos].flag = 0;
}
}
void modify(int l, int r, int pos, int x, int y, int z) {
if (x <= l && y >= r) {
node[pos].w += (ll)(r-l+1) * z;
node[pos].flag += z;
return;
}
push(l, r, pos);
int m = l + r >> 1;
if (x <= m)
modify(l, m, lson(pos), x, y, z);
if (y > m)
modify(m+1, r, rson(pos), x, y, z);
update(pos);
}
ll query(int l, int r, int pos, int x, int y) {
if (x <= l && y >= r)
return node[pos].w;
push(l, r, pos);
int m = l + r >> 1;
ll res = 0;
if (x <= m)
res += query(l, m, lson(pos), x, y);
if (y > m)
res += query(m+1, r, rson(pos), x, y);
return res;
}
} tree;
int main() {
int n, q, a, b, c;
char str[10];
while (scanf("%d%d", &n, &q) != EOF) {
tree.build(1, n, 1);
for (int i = 0; i < q; i++) {
scanf("%s", str);
if (str[0] == 'Q') {
scanf("%d%d", &a, &b);
printf("%lld\n",tree.query(1, n, 1, a, b));
}
else if (str[0] == 'C'){
scanf("%d%d%d", &a, &b, &c);
tree.modify(1, n, 1, a, b, c);
}
}
}
return 0;
}POJ - 3468 A Simple Problem with Integers （区间求和）,布布扣,bubuko.com
POJ - 3468 A Simple Problem with Integers （区间求和）
标签：des style os io for ar div line
原文地址：http://blog.csdn.net/u011345136/article/details/38356063
