标签：区间 member less 技术 ant marked max ide tin
题目链接：https://vjudge.net/problem/HDU-4027
InputThe input contains several test cases, terminated by EOF.
For each test case, the first line contains a single integer N, denoting there are N battleships of evil in a line. (1 <= N <= 100000)
The second line contains N integers Ei, indicating the endurance value of each battleship from the beginning of the line to the end. You can assume that the sum of all endurance value is less than 2 63.
The next line contains an integer M, denoting the number of actions and queries. (1 <= M <= 100000)
For the following M lines, each line contains three integers T, X and Y. The T=0 denoting the action of the secret weapon, which will decrease the endurance value of the battleships between the X-th and Y-th battleship, inclusive. The T=1 denoting the query of the commander which ask for the sum of the endurance value of the battleship between X-th and Y-th, inclusive.
OutputFor each test case, print the case number at the first line. Then print one line for each query. And remember follow a blank line after each test case.Sample Input
10 1 2 3 4 5 6 7 8 9 10 5 0 1 10 1 1 10 1 1 5 0 5 8 1 4 8
Sample Output
Case #1: 19 7 6
代码如下：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <vector> 7 #include <queue> 8 #include <stack> 9 #include <map> 10 #include <string> 11 #include <set> 12 using namespace std; 13 typedef long long LL; 14 const double EPS = 1e-8; 15 const int INF = 2e9; 16 const LL LNF = 2e18; 17 const int MAXN = 1e5+10; 18 19 LL sum[MAXN<<2]; 20 21 void push_up(int u) 22 { 23 sum[u] = sum[u*2] + sum[u*2+1]; 24 } 25 26 void build(int u, int l, int r) 27 { 28 if(l==r) 29 { 30 scanf("%lld", &sum[u]); 31 return; 32 } 33 34 int mid = (l+r)>>1; 35 build(u*2, l, mid); 36 build(u*2+1, mid+1, r); 37 push_up(u); 38 } 39 40 void attack(int u, int l, int r, int x, int y) 41 { 42 if(l==r) 43 { 44 sum[u] = (LL)sqrt(sum[u]); 45 return; 46 } 47 if(x<=l && r<=y && sum[u]==1LL*(r-l+1)) return; 48 49 int mid = (l+r)>>1; 50 if(x<=mid) attack(u*2, l, mid, x, y); 51 if(y>=mid+1) attack(u*2+1, mid+1, r, x, y); 52 push_up(u); 53 } 54 55 LL query(int u, int l, int r, int x, int y) 56 { 57 if(x<=l && r<=y) 58 return sum[u]; 59 60 LL ret = 0; 61 int mid = (l+r)>>1; 62 if(x<=mid) ret += query(u*2, l, mid, x, y); 63 if(y>=mid+1) ret += query(u*2+1, mid+1, r, x, y); 64 return ret; 65 } 66 67 int main() 68 { 69 int n, m, kase = 0; 70 while(scanf("%d", &n)!=EOF) 71 { 72 build(1, 1, n); 73 scanf("%d", &m); 74 printf("Case #%d:\n", ++kase); 75 for(int i = 1; i<=m; i++) 76 { 77 LL op, x, y; 78 scanf("%lld%lld%lld", &op, &x, &y); 79 int xx = min(x, y), yy = max(x, y); 80 if(op==0) attack(1, 1, n, xx, yy ); 81 else printf("%lld\n", query(1, 1, n, xx, yy) ); 82 } 83 printf("\n"); 84 } 85 }
HDU4027 Can you answer these queries? —— 线段树 区间修改
标签：区间 member less 技术 ant marked max ide tin
原文地址：http://www.cnblogs.com/DOLFAMINGO/p/7726053.html
