POJ 3468 线段树
链接：
http://poj.org/problem?id=3468
题解：
就是普通的线段树区间更新，区间求和
代码：
1 #include <map> 2 #include <set> 3 #include <cmath> 4 #include <queue> 5 #include <stack> 6 #include <cstdio> 7 #include <string> 8 #include <vector> 9 #include <cstdlib> 10 #include <cstring> 11 #include <sstream> 12 #include <iostream> 13 #include <algorithm> 14 #include <functional> 15 using namespace std; 16 #define rep(i,a,n) for (int i=a;i<n;i++) 17 #define per(i,a,n) for (int i=n-1;i>=a;i--) 18 #define all(x) (x).begin(),(x).end() 19 #define pb push_back 20 #define mp make_pair 21 #define lson l,m,rt<<1 22 #define rson m+1,r,rt<<1|1 23 typedef long long ll; 24 typedef vector<int> VI; 25 typedef pair<int, int> PII; 26 const ll MOD = 1e9 + 7; 27 const int INF = 0x3f3f3f3f; 28 const int MAXN = 1e5 + 7; 29 // head 30 31 ll n, q; 32 ll tree[MAXN << 2], lazy[MAXN << 2]; 33 34 void pushup(ll rt) { 35 tree[rt] = tree[rt << 1] + tree[rt << 1 | 1]; 36 } 37 38 void pushdown(ll rt, ll m) { 39 if (lazy[rt] != 0) { 40 tree[rt << 1] += lazy[rt] * (m - (m >> 1)); 41 tree[rt << 1 | 1] += lazy[rt] * (m >> 1); 42 lazy[rt << 1] += lazy[rt]; 43 lazy[rt << 1 | 1] += lazy[rt]; 44 lazy[rt] = 0; 45 } 46 } 47 48 void build (ll l, ll r, ll rt) { 49 if (l == r) { 50 scanf("%lld", &tree[rt]); 51 return; 52 } 53 ll m = (l + r) >> 1; 54 build(lson); 55 build(rson); 56 pushup(rt); 57 } 58 59 void update(ll L, ll R, ll x, ll l, ll r, ll rt) { 60 if (L <= l && r <= R) { 61 tree[rt] += (r - l + 1)*x; 62 lazy[rt] += x; 63 return; 64 } 65 pushdown(rt, r - l + 1); 66 ll m = (l + r) >> 1; 67 if (L <= m) update(L, R, x, lson); 68 if (R > m) update(L, R, x, rson); 69 pushup(rt); 70 } 71 72 ll query(ll L, ll R, ll l, ll r, ll rt) { 73 if (L <= l && r <= R) return tree[rt]; 74 pushdown(rt, r - l + 1); 75 ll m = (l + r) >> 1; 76 ll ret = 0; 77 if (L <= m) ret += query(L, R, lson); 78 if (R > m) ret += query(L, R, rson); 79 return ret; 80 } 81 82 int main() { 83 cin >> n >> q; 84 build(1, n, 1); 85 while (q--) { 86 char ch; 87 ll a, b, c; 88 scanf("\n%c", &ch); 89 if (ch == 'C') { 90 scanf("%lld%lld%lld", &a, &b, &c); 91 update(a, b, c, 1, n, 1); 92 } 93 else { 94 scanf("%lld%lld", &a, &b); 95 printf("%lld\n", query(a, b, 1, n, 1)); 96 } 97 } 98 return 0; 99 }
