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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.
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.
You need to answer all Q commands in order. One answer in a line.
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
4
55
9
15
线段树，区间修改，区间求和
CODE:
#include <iostream> #include <cstdio> #include <cstring> #define REP(i, s, n) for(int i = s; i <= n; i ++) #define REP_(i, s, n) for(int i = n; i >= s; i --) #define MAX_N 100000 + 10 using namespace std; int n, q, data[MAX_N]; typedef long long LL; struct segtree{ int l, r; LL sum, c; }a[MAX_N << 2]; void update(int i){ int t1 = i << 1, t2 = t1 + 1; a[i].sum = a[t1].sum + a[t2].sum; } void Make_Tree(int i, int l, int r){ a[i].l = l; a[i].r = r; if(a[i].l == a[i].r){ a[i].sum = data[l]; a[i].c = 0; return; } int mid = (a[i].l + a[i].r) >> 1, t1 = i << 1, t2 = t1 + 1; Make_Tree(t1, l, mid); Make_Tree(t2, mid + 1, r); update(i); } void lazy(int i){ int t1 = i << 1, t2 = t1 + 1; a[t1].sum += (LL)(a[t1].r - a[t1].l + 1) * a[i].c; a[t2].sum += (LL)(a[t2].r - a[t2].l + 1) * a[i].c; a[t1].c += a[i].c; a[t2].c += a[i].c; a[i].c = 0; } void Modify(int i, int l, int r, int c){ if(l > r) return; if(a[i].l == l && a[i].r == r){ a[i].c += c; a[i].sum += (LL)(r - l + 1) * c; return; } lazy(i); int mid = (a[i].l + a[i].r) >> 1, t1 = i << 1, t2 = t1 + 1; if(r <= mid) Modify(t1, l, r, c); else if(l > mid) Modify(t2, l, r, c); else Modify(t1, l, mid, c), Modify(t2, mid + 1, r, c); update(i); } LL Query(int i, int l, int r){ if(l > r) return 0; if(a[i].l == l && a[i].r == r) return a[i].sum; lazy(i); int mid = (a[i].l + a[i].r) >> 1, t1 = i << 1, t2 = t1 + 1; if(r <= mid) return Query(t1, l, r); else if(l > mid) return Query(t2, l, r); else return Query(t1, l, mid) + Query(t2, mid + 1, r); } int main(){ freopen("A.in", "r", stdin); scanf("%d%d", &n, &q); REP(i, 1, n) scanf("%d", &data[i]); Make_Tree(1, 1, n); char s[3]; int x, y, k; while(q --){ scanf("%s", s); if(s[0] == ‘C‘) scanf("%d%d%d", &x, &y, &k), Modify(1, x, y, k); else scanf("%d%d", &x, &y), cout << Query(1, x, y) << "\n"; } return 0; }
POJ 3468 A Simple Problem with Integers
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原文地址：http://www.cnblogs.com/ALXPCUN/p/4608480.html
