A Simple Problem with Integers POJ - 3468
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.
题解：注意add也可能超过int
1 #pragma warning(disable:4996) 2 #include<cstdio> 3 #include<string> 4 #include<cstring> 5 #include<iostream> 6 #include<algorithm> 7 using namespace std; 8 9 #define lson root<<1 10 #define rson root<<1|1 11 #define ll long long 12 13 const int maxn = 100005; 14 15 struct node { 16 int l, r; 17 ll sum, add; 18 }Tree[4 * maxn]; 19 20 void Pushup(int root) { 21 Tree[root].sum = Tree[lson].sum + Tree[rson].sum; 22 } 23 24 void Pushdown(int l, int r, int root) { 25 int mid = (l + r) >> 1; 26 Tree[lson].add += Tree[root].add; 27 Tree[rson].add += Tree[root].add; 28 Tree[lson].sum += Tree[root].add * (mid - l + 1); 29 Tree[rson].sum += Tree[root].add * (r - mid); 30 Tree[root].add = 0; 31 } 32 33 void Build(int l, int r, int root) { 34 Tree[root].l = l; 35 Tree[root].r = r; 36 Tree[root].add = 0; 37 if (l == r) { 38 scanf("%lld", &Tree[root].sum); 39 return; 40 } 41 int mid = (l + r) >> 1; 42 Build(l, mid, lson); 43 Build(mid + 1, r, rson); 44 Pushup(root); 45 } 46 47 void Update(int L, int R, int l, int r, int root, int x) { 48 if (l > R || r < L) return; 49 if (L <= l && r <= R) { 50 Tree[root].add += x; 51 Tree[root].sum += (ll)x * (r - l + 1); 52 return; 53 } 54 if (Tree[root].add) Pushdown(l, r, root); 55 int mid = (l + r) >> 1; 56 Update(L, R, l, mid, lson, x); 57 Update(L, R, mid + 1, r, rson, x); 58 Pushup(root); 59 } 60 61 ll Query(int L, int R, int l, int r, int root) { 62 if (l > R || r < L) return 0; 63 if (L <= l && r <= R) return Tree[root].sum; 64 if (Tree[root].add) Pushdown(l, r, root); 65 int mid = (l + r) >> 1; 66 67 ll ans = 0; 68 ans += Query(L, R, l, mid, lson); 69 ans += Query(L, R, mid + 1, r, rson); 70 return ans; 71 } 72 73 int n, q; 74 75 int main() 76 { 77 while (scanf("%d%d", &n, &q) != EOF) { 78 Build(1, n, 1); 79 char op[10]; 80 for (int i = 1; i <= q; i++) { 81 scanf("%s", op); 82 if (op[0] == 'Q') { 83 int x, y; 84 scanf("%d%d", &x, &y); 85 printf("%lld\n", Query(x, y, 1, n, 1)); 86 } 87 else { 88 int x, y, d; 89 scanf("%d%d%d", &x, &y, &d); 90 Update(x, y, 1, n, 1, d); 91 } 92 } 93 } 94 return 0; 95 }
