题目链接：http://poj.org/problem?id=3468
线段树区间更新查询的样题，注意数据范围。连要更新的数据也必须是long long。
1 #include <algorithm> 2 #include <iostream> 3 #include <iomanip> 4 #include <cstring> 5 #include <climits> 6 #include <complex> 7 #include <cassert> 8 #include <cstdio> 9 #include <bitset> 10 #include <vector> 11 #include <deque> 12 #include <queue> 13 #include <stack> 14 #include <ctime> 15 #include <set> 16 #include <map> 17 #include <cmath> 18 using namespace std; 19 20 typedef long long LL; 21 #define lrt rt << 1 22 #define rrt rt << 1 | 1 23 const int maxn = 100100; 24 LL sum[maxn<<2], add[maxn<<2]; 25 int n, q; 26 char cmd[2]; 27 28 void pushUP(int rt) { 29 sum[rt] = sum[lrt] + sum[rrt]; 30 } 31 32 void pushDOWN(int rt, int len) { 33 if(add[rt]) { 34 sum[lrt] += (len - len / 2) * add[rt]; 35 sum[rrt] += len / 2 * add[rt]; 36 add[lrt] += add[rt]; 37 add[rrt] += add[rt]; 38 add[rt] = 0; 39 } 40 } 41 42 void build(int l, int r, int rt) { 43 add[rt] = 0; 44 if(l == r) { 45 scanf("%I64d", &sum[rt]); 46 return; 47 } 48 int mid = (l + r) >> 1; 49 build(l, mid, lrt); 50 build(mid+1, r, rrt); 51 pushUP(rt); 52 } 53 54 void update(int L, int R, LL c, int l, int r, int rt) { 55 if(L <= l && r <= R) { 56 add[rt] += c; 57 sum[rt] += (LL)c * (r - l + 1); 58 return; 59 } 60 pushDOWN(rt, r-l+1); 61 int mid = (l + r) >> 1; 62 if(L <= mid) update(L, R, c, l, mid, lrt); 63 if(mid < R) update(L, R, c, mid+1, r, rrt); 64 pushUP(rt); 65 } 66 67 LL query(int L, int R, int l, int r, int rt) { 68 if(L <= l && r <= R) return sum[rt]; 69 pushDOWN(rt, r-l+1); 70 int mid = (l + r) >> 1; 71 LL ret = 0; 72 if(L <= mid) ret += query(L, R, l, mid, lrt); 73 if(mid < R) ret += query(L, R, mid+1, r, rrt); 74 return ret; 75 } 76 77 int main() { 78 // freopen("in", "r", stdin); 79 int l, r, c; 80 while(~scanf("%d%d",&n,&q)) { 81 build(1, n, 1); 82 while(q--) { 83 scanf("%s", cmd); 84 if(cmd[0] == 'Q') { 85 scanf("%d%d",&l,&r); 86 printf("%I64d ", query(l,r,1,n,1)); 87 } 88 else { 89 scanf("%d%d%d",&l,&r,&c); 90 update(l,r,LL(c),1,n,1); 91 } 92 } 93 } 94 return 0; 95 }