标签：out body inpu name desc def res pac oid
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 163977 | Accepted: 50540 | |
| 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
#include<iostream> #include<algorithm> #include<string.h> #include<string> #define ll long long using namespace std; ll tree[8000050], lazy[8000050], len[8000050];//tree[num]存的是节点num所在区间的区间和 void pushdown(ll num) { if (lazy[num] != 0) { tree[num * 2] = tree[num * 2] + lazy[num] * len[num * 2]; tree[num * 2 + 1] = tree[num * 2 + 1] + lazy[num] * len[num * 2 + 1]; lazy[num * 2] = lazy[num * 2] + lazy[num]; lazy[num * 2 + 1] = lazy[num * 2 + 1] + lazy[num]; lazy[num] = 0; } } void build(ll num, ll le, ll ri) { len[num] = ri - le + 1;//区间长度 if (le == ri) { scanf("%lld", &tree[num]); return; } ll mid = (le + ri) / 2; build(num * 2, le, mid); build(num * 2 + 1, mid + 1, ri); tree[num] = tree[num * 2] + tree[num * 2 + 1]; } void update(ll num, ll le, ll ri, ll x, ll y, ll z) { if (x <= le && ri <= y) { lazy[num] = lazy[num] + z; tree[num] = tree[num] + len[num] * z;//更新区间和 return; } pushdown(num); ll mid = (le + ri) / 2; if (x <= mid) update(num * 2, le, mid, x, y, z); if (y > mid) update(num * 2 + 1, mid + 1, ri, x, y, z); tree[num]=tree[num*2]+tree[num*2+1]; } ll query(ll num, ll le, ll ri, ll x, ll y) { if (x <= le && ri <= y)//查询区间在num节点所在区间内 return tree[num]; pushdown(num); ll mid = (le + ri) / 2; ll ans = 0; if (x <= mid) ans = ans + query(num * 2, le, mid, x, y); if (y > mid) ans = ans + query(num * 2 + 1, mid + 1, ri, x, y); return ans; } int main() { ll n, m; scanf("%lld%lld", &n, &m); build(1, 1, n); while (m--) { char c[15]; scanf("%s", c); if (c[0] == ‘Q‘) { ll x, y; scanf("%lld%lld", &x, &y); printf("%lld\n", query(1, 1, n, x, y)); } else { ll x, y, z; scanf("%lld%lld%lld", &x, &y, &z); update(1, 1, n, x, y, z); } } return 0; }
POJ 3468 区间更新（求任意区间和）A Simple Problem with Integers
标签：out body inpu name desc def res pac oid
原文地址：https://www.cnblogs.com/-citywall123/p/11266535.html
