POJ A Simple Problem with Integers （线段树区间更新区间查询模板）

题目链接：http://poj.org/problem?id=3468

题意：N个数，Q个操作，操作有两种，‘Q a b ’是询问a~b这段数的和，‘C a b c’是把a~b这段数都加上c。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
struct node{
int l, r, m;
ll val, book;
};
const int maxn = 100100;
node T[maxn << 2];
int a[maxn];
void pushup(int rt) {
T[rt].val = T[rt << 1].val + T[rt << 1 | 1].val;
}
void pushdown(int rt) {
if(T[rt].book) {
T[rt << 1].book += T[rt].book;
T[rt << 1 | 1].book += T[rt].book;
T[rt << 1].val += T[rt].book * (T[rt << 1].r - T[rt << 1].l + 1);
T[rt << 1 | 1].val += T[rt].book * (T[rt << 1 | 1].r - T[rt << 1 | 1].l + 1);
T[rt].book = 0;
}
}
void build(int begin, int end, int rt) {
T[rt].l = begin;
T[rt].r = end;
T[rt].m = (T[rt].l + T[rt].r) >> 1;
T[rt].book = 0;
if(begin == end) {
T[rt].val = a[begin];
return ;
}
build(T[rt].l, T[rt].m, rt << 1);
build(T[rt].m + 1, T[rt].r, rt << 1 | 1);
pushup(rt);
}
void update(int L, int R, int num, int rt) {
if(L <= T[rt].l && T[rt].r <= R) {
T[rt].book += num;
T[rt].val += (ll)num * (T[rt].r - T[rt].l + 1);
return ;
}
pushdown(rt);
if(L > T[rt].m) update(L, R, num, rt << 1 | 1);
else if(R <= T[rt].m) update(L, R, num, rt << 1);
else {
update(L, T[rt].m, num, rt << 1);
update(T[rt].m + 1, R, num, rt << 1 | 1);
}
pushup(rt);
}
ll query(int L, int R, int rt) {
if(L <= T[rt].l && T[rt].r <= R) {
return T[rt].val;
}
pushdown(rt);
if(R <= T[rt].m) return query(L, R, rt << 1);
else if(L > T[rt].m) return query(L, R, rt << 1 | 1);
else return query(L, T[rt].m, rt << 1) + query(T[rt].m + 1, R, rt << 1 | 1);
}

int main() {
int n, q;
scanf("%d %d", &n, &q);
for(int i = 1; i <= n; i++) scanf("%d", a + i);
build(1, n, 1);
getchar();
char op;
int x, y, z;
while(q--) {
scanf("%c", &op);
if(op == 'Q') {
scanf("%d %d", &x, &y);
printf("%I64d\n", query(x, y, 1));
}
else {
scanf("%d %d %d", &x, &y, &z);
update(x, y, z, 1);
}
getchar();
}
return 0;
}