题意：给定一个序列，动态更新值,然后多次询问，求每段和
解题思路：线段树基本操作
解题代码：
1 #include <stdio.h> 2 #include <string.h> 3 #include <stdlib.h> 4 #include <math.h> 5 #define MAX 50005 6 int a[MAX]; 7 int ans = 0 ; 8 int L(int c) 9 { 10 return c * 2; 11 } 12 int R(int c) 13 { 14 return c * 2 + 1; 15 } 16 17 struct node 18 { 19 int left , right ,mid; 20 int sum; 21 }tree[MAX * 3]; 22 void up(int c) 23 { 24 tree[c].sum = tree[L(c)].sum + tree[R(c)].sum; 25 } 26 void build(int c, int p , int v ) 27 { 28 tree[c].left = p ; 29 tree[c].right = v; 30 tree[c].mid = (p + v )/2; 31 tree[c].sum = 0 ; 32 if(p == v ) 33 { 34 tree[c].sum = a[p]; 35 return ; 36 } 37 build(L(c),p,tree[c].mid); 38 build(R(c),tree[c].mid+1,v); 39 up(c); 40 } 41 42 void update(int c, int p, int value) 43 { 44 if(tree[c].left == p && tree[c].right == p) 45 { 46 tree[c].sum += value; 47 return ; 48 } 49 if(p <= tree[c].mid) 50 { 51 update(L(c),p,value); 52 } 53 else 54 update(R(c),p,value); 55 up(c); 56 } 57 58 void sum (int c, int p , int v ) 59 { 60 if(p <= tree[c].left && v >= tree[c].right) 61 { 62 ans += tree[c].sum; 63 return ; 64 } 65 if(tree[c].mid < p ) sum(R(c),p,v); 66 else if(tree[c].mid >= v) sum(L(c),p,v); 67 else 68 { 69 sum(L(c),p,tree[c].mid); 70 sum(R(c),tree[c].mid+1,v); 71 } 72 } 73 int main() 74 { 75 int t ; 76 scanf("%d",&t); 77 for(int CASE = 1; CASE <= t ; CASE ++) 78 { 79 int n ; 80 scanf("%d",&n); 81 for(int i = 1;i <= n;i ++) 82 scanf("%d",&a[i]); 83 build(1,1,n); 84 85 printf("Case %d: ",CASE); 86 for(int i = 1; ;i ++) 87 { 88 char str[10]; 89 scanf("%s",str); 90 91 if(str[0] == 'E') 92 break; 93 else if(str[0] == 'A') 94 { 95 96 int a, b ; 97 scanf("%d %d",&a,&b); 98 update(1,a,b); 99 } 100 else if(str[0] == 'S') 101 { 102 int a, b ; 103 scanf("%d %d",&a,&b); 104 update(1,a,-b); 105 } 106 else if(str[0] == 'Q') 107 { 108 int a, b ; 109 scanf("%d %d",&a,&b); 110 ans = 0 ; 111 sum(1,a,b); 112 printf("%d ",ans); 113 } 114 115 } 116 } 117 return 0 ; 118 }
ps：线段树第一题，学是找了好久的资料啊，过几天再写总结吧