【HDU1166】敌兵布阵（树状数组或线段树）

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。

 #include <iostream>
 #include <cstring>
 #include <cstdlib>
 #include <cstdio>
 #include <numeric>
 #include <algorithm>
 #include <cmath>
 #include <cctype>
 #include <string>
 using namespace std;

 const int N = ;
 int C[N];

 struct peo {
     int x, v;
 }P[N];

 int lowbit (int x) {
     return x & -x;
 }

 int sum (int x) {
     int ret = ;
     while (x > ) {
         ret += C[x]; x -= lowbit(x);
     }
     return ret;
 }

 void add (int x, int d) {
     while (x <= ) {
         C[x] += d; x += lowbit(x);
     }
 }

 int _sum (int d, int u) {
     return sum(u) - sum(d - );
 }

 int main () {
     int T, n, cur = ; scanf("%d", &T);
     char op[];
     while (T --) {
         memset(C, , sizeof(C));
         memset(P, , sizeof(P));
         scanf("%d", &n);
         for (int i =  ; i < n; ++ i) {
             scanf("%d", &P[i].v);
             P[i].x = i + ;
             add (P[i].x, P[i].v);
         }
         //cout << sum(3) << endl;
         printf("Case %d:\n", ++ cur);
         while (scanf("%s", op)) {
             if (op[] == 'E') {
                 break;
             } else if (op[] == 'Q') {
                 int x, y;
                 scanf("%d %d", &x, &y);
                 cout << _sum(x, y) << endl;
             } else if (op[] == 'S') {
                 int x, y;
                 scanf("%d %d", &x, &y);
                 add (x, -y);
             } else if (op[] == 'A') {
                 int x, y;
                 scanf("%d %d", &x, &y);
                 add (x, y);
             }
         }
     }
     return ;
 }

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <vector>
 #include <cstring>
 #include <cstdlib>
 #include <cctype>
 #include <algorithm>
 #include <numeric>

 using namespace std;

 struct P{
     int sum, l, r;
 } node[];

 int n, num[] = {};

 void build (int k, int l, int r) {
     node[k].l = l, node[k].r = r;
     if (l == r) {
         node[k].sum = num[l];
     } else {
         build( * k, l, (l + r) / );
         build( * k + , ((l + r) / ) + , r);
         node[k].sum = node[ * k].sum + node[ * k + ].sum;
     }
 }

 void update(int k, int i, int x) {
     if (node[k].l <= i && node[k].r >= i) {
         node[k].sum += x;
     }
     if (node[k].l == node[k].r) {
         return ;
     }
     if (i <= node[ * k].r) {
         update( * k, i, x);
     }
     else update( * k + , i, x);
 }

 int query (int k, int l, int r) {
     if (l == node[k].l && r == node[k].r) {
         return node[k].sum;
     }
     if (r <= node[ * k].r)  {
         return query( * k, l, r);
     }
     if (l >= node[ * k + ].l) {
         return query( * k + , l, r);
     } else {
         return query( * k, l, node[ * k].r) + query( * k + , node[ * k + ].l, r);
     }
 }

 int main(){
     int T;
     scanf("%d",&T);
     for (int z = ; z <= T; ++ z) {
         memset(num, , sizeof(num));
         scanf("%d", &n);
         for (int i = ; i <= n; ++ i) {
             scanf("%d",&num[i]);
         }
         build(, , n);
         char s[];
         printf("Case %d:\n",z);
         while (scanf("%s", s)){
             if (s[] == 'E')    break;
             if (s[] == 'Q'){
                 int a, b;
                 scanf("%d%d" ,&a ,&b);
                 printf("%d\n",query(, a, b));
             }
             if (s[] == 'A'){
                 int a,x;
                 scanf("%d%d", &a, &x);
                 update(, a, x);
             }
             if (s[] == 'S') {
                 int a,x;
                 scanf("%d%d", &a, &x);
                 update(, a, -x);
             }
         }
     }
     return ;
 }