POJ 3468 A Simple Problem with Integers
线段树，维护区间的和，支持区间范围修改，注意由于增加的标记传递时是累加起来，结果会超出 int ，要用 long long；
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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
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# include <stdio.h> # define N 100005 # define NON 0 typedef long long int LL; int n, q, num[N]; LL sum[4 * N], ans; LL bj[4 * N]; void update(int r) { sum[r] = sum[r << 1] + sum[(r << 1) | 1]; } void build(int r, int x, int y) { bj[r] = NON; if (x == y) { sum[r] = num[x]; return ; } int mid = (x+y) >> 1, ls = r << 1, rs = (r << 1) | 1; build(ls, x, mid); build(rs, mid+1, y); update(r); } void pushdown(int r, int x, int y) { if (bj[r] != NON) { int mid = (x+y) >> 1, ls = r << 1, rs = (r << 1) | 1; sum[ls] += (mid-x+1) * bj[r]; sum[rs] += (y - mid) * bj[r]; bj[ls] += bj[r], bj[rs] += bj[r], bj[r] = NON; } } void add(int r, int x, int y, int s, int t, int val) { if (s<=x && y<=t) { bj[r] += val; sum[r] += (y-x+1) * val; return ; } pushdown(r, x, y); int mid = (x+y) >> 1, ls = r << 1, rs = (r << 1) | 1; if (s <= mid) add(ls, x, mid, s, t, val); if (mid+1 <= t) add(rs, mid+1, y, s, t, val); update(r); } void query(int r, int x, int y, int s, int t, LL &ans) { if (s<=x && y<=t) { ans += sum[r]; return ; } pushdown(r, x, y); int mid = (x+y) >> 1, ls = r << 1, rs = (r << 1) | 1; if (s <= mid) query(ls, x, mid, s, t, ans); if (mid+1 <= t) query(rs, mid+1, y, s, t, ans); } void init(void) { for (int i = 1; i <= n; ++i) scanf("%d", &num[i]); build(1, 1, n); } void solve(void) { int s, t, val; char str[5]; for (int i = 1; i <= q; ++i) { scanf("%s", str); switch(str[0]) { case 'C': {scanf("%d%d%d", &s, &t, &val); add(1, 1, n, s, t, val); break;} case 'Q': {scanf("%d%d", &s, &t); ans = 0; query(1, 1, n, s, t, ans); printf("%lld\n", ans); break;} } } } int main() { while (~scanf("%d%d", &n, &q)) { init(); solve(); } return 0; }
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