POJ 3468 线段树 区间修改 基础题

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 87262		Accepted: 27084
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

The sums may exceed the range of 32-bit integers.

题目大意：

给一个序列，C是修改，Q是求值。

恩。。。sum[o]在maintain中的时候要重新赋值，不能直接用书上的sum[o] = 0，而是要之前保存一个sum的值的数组，然后用sum[o] = a[o]才可以。

其他方面好像就没有什么了吧

#include
#include
#include

using namespace std;

typedef long long ll;
const int maxn = 100000 + 5;
ll sum[maxn * 4], add[maxn * 4];
ll a[maxn * 4];
ll n, q;
ll ans;

void buildtree(int o, int l, int r){
if (l == r){
scanf("%I64d", &sum[o]);
a[o] = sum[o];
return ;
}
int mid = l + (r - l) / 2;
buildtree(o << 1, l, mid);
buildtree(o << 1 | 1, mid + 1, r);
sum[o] = sum[o << 1] + sum[o << 1 | 1];
a[o] = sum[o];
}

void init(){
memset(sum, 0, sizeof(sum));
memset(add, 0, sizeof(add));
memset(a, 0, sizeof(a));
buildtree(1, 1, n);
}

void query(int o, int ql, int qr, int l, int r, ll plu){
if (ql <= l && qr >= r){
ans += sum[o] + plu * (r - l + 1);
}
else {
int mid = l + (r - l) / 2;
if (mid >= ql){
query(o << 1, ql, qr, l, mid, plu + add[o]);
}
if(mid < qr){
query(o << 1 | 1, ql, qr, mid + 1, r, plu + add[o]);
}
}

}

void maintain(int o, int l, int r){
sum[o] = a[o];
if (l < r){
sum[o] = sum[o << 1] + sum[o << 1 | 1];
}
sum[o] += add[o] * (r - l + 1);
return ;
}

void update(int o, int ql, int qr, ll plu, int l, int r){
if (ql <= l && qr >= r){
add[o] += plu;
}
else {
int mid = l + (r - l) / 2;
if (mid >= ql){
update(o << 1, ql, qr, plu, l, mid);
}
if (mid < qr){
update(o << 1 | 1, ql, qr, plu, mid + 1, r);
}
}
maintain(o, l, r);
}

void solve(){
while (q--){
char ch[2];
scanf("%s", ch);
if (ch[0] == 'Q'){
ans = 0;
int l, r;
scanf("%d%d", &l, &r);
query(1, l, r, 1, n, 0);
printf("%I64d\n", ans);
}
else {
int l, r;
ll plu;
scanf("%d%d%I64d", &l, &r, &plu);
update(1, l, r, plu, 1, n);
}
}
}

int main(){
while (scanf("%I64d%I64d", &n, &q) == 2){
init();
solve();
}
return 0;
}