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
【分析】
以前一直以为树状数组只能单点修改，区间询问，今天算是见识到了树状数组的区间修改了。
不说多的了，很巧妙的感觉，对于一个修改操作(a,b,c)，将[a,b]区间整体加上一个c
可以把它拆开成两个部分(a,n,c),和(b + 1,n,-c)，这个时候就好做了，用两个树状数组。
一个树状数组记录从修改点开始一直到结束所增加的所有量，及c*(n-a+1)。另一个记录下增量值c。
那么对于一个在a点右边而在b点左边的点x,它所得到的增量为c*(n-a+1) - c*(n-x)。很好维护了。
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <cstring> 5 #include <vector> 6 #include <utility> 7 #include <iomanip> 8 #include <string> 9 #include <cmath> 10 #include <map> 11 12 const int MAXN = 100000 + 10; 13 const int MAX = 32000 + 10; 14 using namespace std; 15 typedef long long ll; 16 int n, m; 17 ll sum[MAXN];//记录原始序列 18 ll C[2][MAXN]; 19 20 int lowbit(int x){return x&-x;} 21 ll get(int k, int x){ 22 ll cnt = 0; 23 while (x > 0){ 24 cnt += C[k][x]; 25 x -= lowbit(x); 26 } 27 return cnt; 28 } 29 void add(int k, int x, ll val){ 30 while (x <= n){ 31 C[k][x] += val; 32 x += lowbit(x); 33 } 34 return ; 35 } 36 void work(){ 37 memset(C, 0, sizeof(C));//0记录总值、1录增量 38 sum[0] = 0; 39 scanf("%d%d", &n, &m); 40 for (int i = 1; i <= n; i++){ 41 scanf("%lld", &sum[i]); 42 sum[i] += sum[i - 1]; 43 } 44 for (int i = 1; i <= m; i++){ 45 char str[2]; 46 scanf("%s", str); 47 if (str[0] == 'Q'){ 48 int l, r; 49 scanf("%d%d", &l, &r); 50 printf("%lld ", sum[r] - sum[l - 1] +(get(0, r) - get(1, r) * (n - r)) - (get(0, l - 1) - get(1, l - 1) * (n - l + 1))); 51 }else{ 52 int l, r; 53 ll x; 54 scanf("%d%d%lld", &l, &r, &x); 55 add(0, l, (n - l + 1) * x); 56 add(0, r + 1, (n - r) * (-x)); 57 add(1, l, x); 58 add(1, r + 1, -x); 59 } 60 } 61 //printf(" %d ", get(0, 10)); 62 } 63 64 int main(){ 65 int T; 66 #ifdef LOCAL 67 freopen("data.txt", "r", stdin); 68 freopen("out.txt", "w", stdout); 69 #endif 70 work(); 71 return 0; 72 }