A Simple Problem with Integers
Time Limit:5000MS Memory Limit:131072KB 64bit IO Format:%I64d & %I64u
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.
1 //Memory: 4324 KB Time: 1594 MS 2 //Language: C++ Result: Accepted 3 4 #include <iostream> 5 #include <cstdio> 6 7 using namespace std; 8 9 typedef long long ll; 10 #define ls rt<<1 11 #define rs rt<<1|1 12 13 const int sz = 100001; 14 struct Tree 15 { 16 ll sum, inc; 17 }t[sz<<2]; 18 ll d, ans; 19 20 void build(const int left, const int right, const int rt) 21 { 22 t[rt].inc = 0; 23 if(right == left) 24 { 25 scanf("%I64d", &t[rt].sum); 26 return ; 27 } 28 int mid = (right + left) >> 1; 29 build(left, mid, ls); 30 build(mid + 1, right, rs); 31 t[rt].sum = t[ls].sum + t[rs].sum; 32 return ; 33 } 34 35 void update(const int l, const int r, const int from, const int to, const int rt) 36 { 37 if(from <= l && r <= to) 38 { 39 t[rt].inc += d; 40 t[rt].sum += ((ll)(r - l + 1) * d); //（1） 41 return ; 42 } 43 if(t[rt].inc) 44 { 45 t[ls].inc += t[rt].inc; 46 t[rs].inc += t[rt].inc; 47 t[ls].sum += ((ll)((r - l + 2) >> 1) * t[rt].inc); //(2) 48 t[rs].sum += ((ll)((r - l + 1) >> 1) * t[rt].inc); 49 t[rt].inc = 0; 50 } 51 int mid = (l + r) >> 1; 52 if(from <= mid) update(l, mid, from, to, ls); 53 if(to > mid) update(mid + 1, r, from, to, rs); 54 t[rt].sum = t[ls].sum + t[rs].sum; 55 return ; 56 } 57 58 void query(const int l, const int r, const int from, const int to, const int rt) 59 { 60 if(from <= l && r <= to) 61 { 62 ans += t[rt].sum; 63 return ; 64 } 65 if(t[rt].inc) 66 { 67 t[ls].inc += t[rt].inc; 68 t[rs].inc += t[rt].inc; 69 t[ls].sum += ((ll)((r - l + 2) >> 1) * t[rt].inc); 70 t[rs].sum += ((ll)((r - l + 1) >> 1) * t[rt].inc); 71 t[rt].inc = 0; 72 } 73 int mid = (l + r) >> 1; 74 if(from <= mid) query(l, mid, from, to, ls); 75 if(to > mid) query(mid + 1, r, from, to, rs); 76 return ; 77 } 78 79 int main() 80 { 81 int q, n, a, b; 82 char cmd; 83 while(scanf("%d %d", &n, &q) != EOF) 84 { 85 build(1, n, 1); 86 while(q--) 87 { 88 cin >> cmd; 89 if(cmd == 'Q') 90 { 91 scanf("%d %d", &a, &b); 92 ans = 0; 93 query(1, n, a, b, 1); 94 printf("%I64d\n", ans); 95 } 96 else 97 { 98 scanf("%d %d %I64d", &a, &b, &d); 99 update(1, n, a, b, 1); 100 } 101 } 102 } 103 return 0; 104 } 105 /* 106 (1) *d而非*t[rt].inc，因为t[rt].inc是保存rt的儿子的更新数据的，到达rt的更新是d 107 (2) *t[rt].inc而非*t[ls].inc，因为t[ls].inc是保存ls的儿子的更新数据的。到达ls的更新是 108 t[rt].inc.(r - l + 2) >> 1是ls的左儿子的区间长度，(r - l + 1) >> 1是右儿子的区间长度。 109 实际上(r - l + 2) >> 1是(r - l + 1) / 2.0的向上取整。注意由于在build中，mid归于左儿子， 110 故左儿子的区间长度>=右儿子区间长度 111 */