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poj 3468线段树区间更新
阅读量:4205 次
发布时间:2019-05-26

本文共 2859 字,大约阅读时间需要 9 分钟。

A Simple Problem with Integers
Time Limit: 5000MS   Memory Limit: 131072K
Total Submissions: 104022   Accepted: 32496
Case Time Limit: 2000MS

Description

You have N integers, A1A2, ... , 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 A1A2, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of AaAa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of AaAa+1, ... , Ab.

Output

You need to answer all Q commands in order. One answer in a line.

Sample Input

10 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4

Sample Output

455915
#include 
#include
#include
#define ll long long #define lson l, mid, rt << 1 #define rson mid + 1, r, rt << 1 | 1 int const MAX = 1e5 + 5; ll sum[MAX << 2], add[MAX << 2]; void PushUp(int rt) { sum[rt] = sum[rt << 1] + sum[rt << 1 | 1]; } void PushDown(int ln, int rn, int rt) { if(add[rt]) { sum[rt << 1] += (ll)ln * add[rt]; sum[rt << 1 | 1] += (ll)rn * add[rt]; add[rt << 1] += add[rt]; add[rt << 1 | 1] += add[rt]; add[rt] = 0; } return; } void Build(int l, int r, int rt) { add[rt] = 0; if(l == r) { scanf("%lld", &sum[rt]); return; } int mid = (l + r) >> 1; Build(lson); Build(rson); PushUp(rt); return; } void Update(int L, int R, int c, int l, int r, int rt) { if(L <= l && r <= R) { sum[rt] += (r - l + 1) * c; add[rt] += c; return; } int mid = (l + r) >> 1; PushDown(mid - l + 1, r - mid, rt); if(L <= mid) Update(L, R, c, lson); if(R > mid) Update(L, R, c, rson); PushUp(rt); return; } ll Query(int L, int R, int l, int r, int rt) { if(L <= l && r <= R) return sum[rt]; int mid = (l + r) >> 1; PushDown(mid - l + 1, r - mid, rt); ll ans = 0; if(L <= mid) ans += Query(L, R, lson); if(R > mid) ans += Query(L, R, rson); return ans; } int main() { int n, q; scanf("%d %d", &n, &q); Build(1, n, 1); while(q --) { char s[2]; scanf("%s", s); if(s[0] == 'Q') { int l, r; scanf("%d %d", &l, &r); printf("%lld\n", Query(l, r, 1, n, 1)); } else { int l, r, c; scanf("%d %d %d", &l, &r, &c); Update(l, r, c, 1, n, 1); } } }

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