Time Limit: 5000MS | Memory Limit: 131072K | |
Total Submissions: 130450 | Accepted: 40489 | |
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
Source
基本的线段树练习题目。。注意每次记录区间的累加值,就可以了,注意清零。
代码:
#include<iostream> #include<cstdio> #include<cstring> #define ll long long using namespace std; const int maxn = 1e5+7; int n, q; ll s[maxn]; struct node { int l, r; ll sumseg, Inc; }segTree[maxn*4]; void Build(int i, int l, int r) { segTree[i].l = l; segTree[i].r = r; segTree[i].Inc = 0; if(l == r) { segTree[i].sumseg = s[l]; return ; } int mid = (l+r)>>1; Build(i<<1, l, mid); Build(i<<1|1, mid+1, r); segTree[i].sumseg = segTree[i<<1].sumseg + segTree[i<<1|1].sumseg; //左子树和右子树; } void ADD(int i, int l, int r, ll c) { if(segTree[i].l == l&&segTree[i].r == r) { segTree[i].Inc += c; return ; } segTree[i].sumseg += c*(r - l + 1); int mid = (segTree[i].l + segTree[i].r)>>1; if(r <= mid) ADD(i<<1, l, r, c); else if(l>mid) ADD(i<<1|1, l, r, c); else { ADD(i<<1, l, mid, c); ADD(i<<1|1, mid+1, r, c); } } ll Query(int i, int l, int r) { if(segTree[i].l == l&&segTree[i].r == r) { return segTree[i].sumseg + segTree[i].Inc*(r - l + 1); } segTree[i].sumseg += (segTree[i].r - segTree[i].l + 1)*segTree[i].Inc; int mid = (segTree[i].l + segTree[i].r)>>1; ADD(i<<1, segTree[i].l, mid, segTree[i].Inc); ADD(i<<1|1, mid+1, segTree[i].r, segTree[i].Inc); segTree[i].Inc = 0; if(r <= mid) return Query(i<<1, l, r); else if(l>mid) return Query(i<<1|1, l, r); else return Query(i<<1, l, mid)+Query(i<<1|1, mid+1, r); } void checkc() { int a, b; ll c; scanf("%d%d%lld",&a, &b, &c); ADD(1, a, b, c); } void checkq() { int a, b; scanf("%d%d",&a, &b); printf("%lld\n", Query(1, a, b)); } int main() { int i, j; //freopen("in.txt", "r", stdin); while(~scanf("%d%d",&n,&q)) { for(i = 1; i <= n; i++) scanf("%lld",&s[i]); Build(1, 1, n); for(i = 0; i < q; i++) { char op; cin >> op; if(op == 'C') checkc(); else checkq(); } } return 0; }