挺简单的线段树， 单点更新。（貌似有人没用线段树也过了，不过效率略低）
做完后粘了别人的代码又去试了试，比我的快了50ms左右。
不过有些人把加减看做两个运算确实很浪费。一个Add解决了。
1 #include<stdio.h> 2 #define mid(a,b) ((a+b)>>1) 3 const int N = 50008; 4 5 struct campsite{ 6 7 int left, right; 8 int sum; 9 10 }army[4*N]; 11 12 void build(int root, int l, int r){ 13 14 army[root].left = l; 15 army[root].right = r; 16 army[root].sum = 0; 17 18 if(l == r) 19 return; 20 build(root*2, l, mid(l, r)); 21 build(root*2 + 1, mid(l, r) + 1, r); 22 } 23 24 void minsert(int root, int pose, int value){ 25 26 if(army[root].left == pose && army[root].right == pose){ 27 army[root].sum += value; 28 return; 29 } 30 31 army[root].sum += value; 32 33 if(pose <= mid(army[root].left, army[root].right)) 34 minsert(root*2, pose, value); 35 else 36 minsert(root*2+1, pose, value); 37 } 38 39 void add(int root, int pose, int value){ 40 41 if(army[root].left == pose && army[root].right == pose){ 42 army[root].sum += value; 43 return; 44 } 45 46 army[root].sum += value; 47 48 if(pose <= mid(army[root].left, army[root].right)) 49 add(root*2, pose, value); 50 else 51 add(root*2+1, pose, value); 52 } 53 54 int query(int root, int l, int r){ 55 56 if(army[root].left == l && army[root].right == r) 57 return army[root].sum; 58 59 if(r <= mid(army[root].left, army[root].right)) 60 return query(root*2, l, r); 61 else if(l > mid(army[root].left, army[root].right)) 62 return query(root*2+1, l, r); 63 else{ 64 return query(root*2, l, mid(army[root].left, army[root].right)) 65 + query(root*2+1, mid(army[root].left, army[root].right) + 1, r); 66 } 67 68 } 69 70 /*int main(){ 71 72 build(1, 1, 5); 73 for(int i = 0; i < 15; i++) 74 printf("%d %d ", army[i].left, army[i].right); 75 76 return 0; 77 78 }*/ 79 int main(){ 80 81 int t, nt, a, b; 82 char op[10]; 83 scanf("%d", &t); 84 nt = t; 85 86 while(t --){ 87 int num, hei; 88 scanf("%d",&num); 89 build(1, 1, num); 90 for(int i = 1; i <= num; i++){ 91 scanf("%d", &hei); 92 minsert(1, i, hei); 93 } 94 int flag = 1; 95 while(~scanf("%s", op)){ 96 97 if(op[0] == 'E') 98 break; 99 if(op[0] == 'A'){ 100 scanf("%d %d", &a, &b); 101 add(1, a, b); 102 } 103 if(op[0] == 'S'){ 104 scanf("%d %d", &a, &b); 105 add(1, a, -b); 106 } 107 if(op[0] == 'Q'){ 108 scanf("%d %d", &a, &b); 109 if(flag){ 110 printf("Case %d: ", nt - t); 111 flag = 0; 112 } 113 printf("%d ",query(1, a, b)); 114 } 115 116 } 117 118 } 119 120 return 0; 121 122 }
开始追求一点效率吧