标签：oid printf cst exce code poj3468 names nts lazy
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.
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.
You need to answer all Q commands in order. One answer in a line.
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
4 55 9 15
1 //2017-05-17 2 #include <cstdio> 3 #include <cstring> 4 #include <iostream> 5 #include <algorithm> 6 #define ll long long 7 #define mid ((st[id].l+st[id].r)>>1) 8 #define lson (id<<1) 9 #define rson ((id<<1)|1) 10 11 using namespace std; 12 13 const ll N = 100005; 14 ll arr[N]; 15 struct Node{ 16 ll l, r, sum, lazy; 17 }st[N<<2]; 18 19 void build(int id, int l, int r) 20 { 21 st[id].l = l; st[id].r = r; st[id].lazy = 0; 22 if(l == r){ 23 st[id].sum = arr[l]; 24 return; 25 } 26 build(lson, l, mid); 27 build(rson, mid+1, r); 28 st[id].sum = st[lson].sum+st[rson].sum; 29 } 30 31 void push_down(int id) 32 { 33 if(st[id].lazy != 0){ 34 st[lson].lazy += st[id].lazy; 35 st[rson].lazy += st[id].lazy; 36 st[id].sum += (st[id].r-st[id].l+1)*st[id].lazy; 37 st[id].lazy = 0; 38 } 39 return; 40 } 41 42 ll query(int id, int l, int r) 43 { 44 if(st[id].l == l && st[id].r == r)return st[id].sum+(r-l+1)*st[id].lazy; 45 push_down(id); 46 if(r <= mid)return query(lson, l, r); 47 else if(l > mid)return query(rson, l, r); 48 else return query(lson, l, mid)+query(rson, mid+1, r); 49 } 50 51 void update(int id, int l, int r, int w) 52 { 53 if(st[id].l == l && st[id].r == r){ 54 st[id].lazy += w; 55 return; 56 } 57 st[id].sum += (r-l+1)*w; 58 if(r <= mid)update(lson, l, r, w); 59 else if(l > mid)update(rson, l, r, w); 60 else{ 61 update(lson, l, mid, w); 62 update(rson, mid+1, r, w); 63 } 64 } 65 66 int main() 67 { 68 ll n, q; 69 while(scanf("%lld%lld", &n, &q)!=EOF){ 70 for(int i = 1; i <= n; i++) 71 scanf("%lld", &arr[i]); 72 build(1, 1, n); 73 char op[5]; 74 ll a, b, c; 75 while(q--){ 76 scanf("%s", op); 77 if(op[0] == ‘Q‘){ 78 scanf("%lld%lld", &a, &b); 79 printf("%lld\n", query(1, a, b)); 80 }else if(op[0] == ‘C‘){ 81 scanf("%lld%lld%lld", &a, &b, &c); 82 update(1, a, b, c); 83 } 84 } 85 } 86 87 return 0; 88 }
标签：oid printf cst exce code poj3468 names nts lazy
原文地址：http://www.cnblogs.com/Penn000/p/6869708.html
