是一道树状数组的裸题，也可以说是线段树的对于单点维护的裸题。多做这种题目可以提高自己对基础知识的理解程度，很经典。
1 #include <iostream> 2 #include <cstring> 3 #include <cstdlib> 4 #include <cstdio> 5 #include <numeric> 6 #include <algorithm> 7 #include <cmath> 8 #include <cctype> 9 #include <string> 10 using namespace std; 11 12 13 const int N = 50005; 14 int C[N]; 15 16 struct peo { 17 int x, v; 18 }P[N]; 19 20 int lowbit (int x) { 21 return x & -x; 22 } 23 24 int sum (int x) { 25 int ret = 0; 26 while (x > 0) { 27 ret += C[x]; x -= lowbit(x); 28 } 29 return ret; 30 } 31 32 void add (int x, int d) { 33 while (x <= 50000) { 34 C[x] += d; x += lowbit(x); 35 } 36 } 37 38 int _sum (int d, int u) { 39 return sum(u) - sum(d - 1); 40 } 41 42 int main () { 43 int T, n, cur = 0; scanf("%d", &T); 44 char op[10]; 45 while (T --) { 46 memset(C, 0, sizeof(C)); 47 memset(P, 0, sizeof(P)); 48 scanf("%d", &n); 49 for (int i = 0 ; i < n; ++ i) { 50 scanf("%d", &P[i].v); 51 P[i].x = i + 1; 52 add (P[i].x, P[i].v); 53 } 54 //cout << sum(3) << endl; 55 printf("Case %d: ", ++ cur); 56 while (scanf("%s", op)) { 57 if (op[0] == 'E') { 58 break; 59 } else if (op[0] == 'Q') { 60 int x, y; 61 scanf("%d %d", &x, &y); 62 cout << _sum(x, y) << endl; 63 } else if (op[0] == 'S') { 64 int x, y; 65 scanf("%d %d", &x, &y); 66 add (x, -y); 67 } else if (op[0] == 'A') { 68 int x, y; 69 scanf("%d %d", &x, &y); 70 add (x, y); 71 } 72 } 73 } 74 return 0; 75 }
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <vector> 5 #include <cstring> 6 #include <cstdlib> 7 #include <cctype> 8 #include <algorithm> 9 #include <numeric> 10 11 using namespace std; 12 13 struct P{ 14 int sum, l, r; 15 } node[150050]; 16 17 int n, num[50050] = {0}; 18 19 void build (int k, int l, int r) { 20 node[k].l = l, node[k].r = r; 21 if (l == r) { 22 node[k].sum = num[l]; 23 } else { 24 build(2 * k, l, (l + r) / 2); 25 build(2 * k + 1, ((l + r) / 2) + 1, r); 26 node[k].sum = node[2 * k].sum + node[2 * k + 1].sum; 27 } 28 } 29 30 void update(int k, int i, int x) { 31 if (node[k].l <= i && node[k].r >= i) { 32 node[k].sum += x; 33 } 34 if (node[k].l == node[k].r) { 35 return ; 36 } 37 if (i <= node[2 * k].r) { 38 update(2 * k, i, x); 39 } 40 else update(2 * k + 1, i, x); 41 } 42 43 int query (int k, int l, int r) { 44 if (l == node[k].l && r == node[k].r) { 45 return node[k].sum; 46 } 47 if (r <= node[2 * k].r) { 48 return query(2 * k, l, r); 49 } 50 if (l >= node[2 * k + 1].l) { 51 return query(2 * k + 1, l, r); 52 } else { 53 return query(2 * k, l, node[2 * k].r) + query(2 * k + 1, node[2 * k + 1].l, r); 54 } 55 } 56 57 int main(){ 58 int T; 59 scanf("%d",&T); 60 for (int z = 1; z <= T; ++ z) { 61 memset(num, 0, sizeof(num)); 62 scanf("%d", &n); 63 for (int i = 1; i <= n; ++ i) { 64 scanf("%d",&num[i]); 65 } 66 build(1, 1, n); 67 char s[100]; 68 printf("Case %d: ",z); 69 while (scanf("%s", s)){ 70 if (s[0] == 'E') break; 71 if (s[0] == 'Q'){ 72 int a, b; 73 scanf("%d%d" ,&a ,&b); 74 printf("%d ",query(1, a, b)); 75 } 76 if (s[0] == 'A'){ 77 int a,x; 78 scanf("%d%d", &a, &x); 79 update(1, a, x); 80 } 81 if (s[0] == 'S') { 82 int a,x; 83 scanf("%d%d", &a, &x); 84 update(1, a, -x); 85 } 86 } 87 } 88 return 0; 89 }