A Simple Problem with Integers
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 126311 | Accepted: 39238 | |
| 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
The sums may exceed the range of 32-bit integers.
Source
POJ Monthly--2007.11.25, Yang Yi
1 #include <iostream> 2 #include <cstring> 3 #include <string> 4 #include <algorithm> 5 typedef long long ll; 6 using namespace std; 7 int n, q; 8 struct node 9 { 10 int l, r; 11 ll date;//区间内每个数都要加的数 12 ll datb;//数据的和 13 }a[400005]; 14 15 16 void build(int l,int r,int k) 17 { 18 a[k].l = l; a[k].r = r; 19 if (r == l) 20 { 21 scanf("%lld", &a[k].datb); 22 return; 23 } 24 int mid = (l + r) / 2;//传的是l和r，不是a[k].l和a[k].r 25 build(l, mid , 2 * k); 26 build(mid + 1, r, 2 * k + 1); 27 a[k].datb = a[2 * k].datb + a[2 * k + 1].datb; 28 } 29 30 void add(int k,int l,int r,int v) 31 { 32 if (a[k].l >= l && r >= a[k].r)//如果区间刚好在a[k].l和a[k].r之间 33 a[k].date += v;//每个数都要加，所以放到date中 34 else if(a[k].l<=r&&l<=a[k].r)//部分重叠的话 35 { 36 a[k].datb+= (min(r, a[k].r) - max(l, a[k].l)+1)*v;//我加号居然忘了 37 add(2 * k, l,r, v); 38 add(2 * k + 1,l, r, v); 39 } 40 //否则跳出循环 41 } 42 43 ll Sum(int l, int r,int k) 44 { 45 if (l > a[k].r||r < a[k].l) 46 return 0; 47 else if (l <= a[k].l&&r >= a[k].r) 48 { 49 /*cout << "a[k].l " << a[k].l << " a[k].r " << a[k].r << endl; 50 cout << a[k].datb + a[k].date *(a[k].r - a[k].l + 1) << endl;*/ 51 return a[k].datb + a[k].date *(a[k].r - a[k].l + 1); 52 } 53 else 54 { 55 ll ans = (min(a[k].r, r) - max(a[k].l, l)+1)*a[k].date; 56 ans += Sum(l, r, k * 2);//传的是l和r，不是a[k].l和a[k].r 57 ans += Sum(l, r, k * 2 + 1); 58 return ans; 59 } 60 } 61 62 int main() 63 { 64 while (~scanf("%d%d", &n, &q)) 65 { 66 for (int i = 0; i <= 400000; i++) 67 a[i].datb = a[i].date = 0; 68 build(1, n, 1); 69 while (q--) 70 { 71 char s; 72 int x, y; 73 cin >> s >> x >> y; 74 if (s == 'C') 75 { 76 int z; 77 cin >> z; 78 add(1, x, y, z); 79 } 80 else 81 { 82 ll sum = Sum(x, y, 1); 83 printf("%lld ", sum); 84 } 85 } 86 } 87 return 0; 88 }