POJ3468 A Simple Problem with Integers(线段树)

Description

You have N integers, A₁, A₂, ... , A_N. 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, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a+1, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a, A_a+1, ... , A_b.

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

线段树的成段更新 继续板子

AC代码：

#include "iostream"#include "cstdio"#include "cstring"#include "algorithm"using namespace std;#define lson l, m, rt << 1#define rson m + 1, r, rt << 1 | 1const int maxn = 100010;typedef long long ll;ll add[maxn << 2], sum[maxn << 2];void pushup(int rt){sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];}void pushdown(int rt, int m){if(add[rt]) {add[rt << 1] += add[rt];add[rt << 1 | 1] += add[rt];sum[rt << 1] += add[rt] * (m - (m >> 1));sum[rt << 1 | 1] += add[rt] * (m >> 1);add[rt] = 0;}}void build(int l, int r, int rt){add[rt] = 0;if(l == r) {scanf("%lld", &sum[rt]);return ;}int m = (l + r) >> 1;build(lson);build(rson);pushup(rt);}void update(int L,int R,int c,int l,int r,int rt) {      if (L <= l && r <= R) {          add[rt] += c;          sum[rt] += (ll)c * (r - l + 1);          return ;      }      pushdown(rt , r - l + 1);      int m = (l + r) >> 1;      if (L <= m) update(L , R , c , lson);      if (m < R) update(L , R , c , rson);      pushup(rt);  } ll query(int L, int R, int l, int r, int rt){if(L <= l && r <= R) return sum[rt];pushdown(rt, r - l + 1);int m = (l + r) >> 1;ll ret = 0;if(L <= m) ret += query(L, R, lson);if(m < R) ret += query(L, R, rson);return ret;}int main(int argc, char const *argv[]){int n, t;scanf("%d%d", &n, &t);build(1, n, 1);while(t--) {char op[2];int a, b, c;scanf("%s", op);if(op[0] == 'Q') {scanf("%d%d", &a, &b);printf("%lld\n", query(a, b, 1, n, 1));}else {scanf("%d%d%d", &a, &b, &c);update(a, b, c, 1, n, 1);}}return 0;}