【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 = 50005;
int C[N];

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

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

int sum (int x) {
    int ret = 0;
    while (x > 0) {
        ret += C[x]; x -= lowbit(x);
    }
    return ret;
}

void add (int x, int d) {
    while (x <= 50000) {
        C[x] += d; x += lowbit(x);
    }
}

int _sum (int d, int u) {
    return sum(u) - sum(d - 1);
}

int main () {
    int T, n, cur = 0; scanf("%d", &T);
    char op[10];
    while (T --) {
        memset(C, 0, sizeof(C));
        memset(P, 0, sizeof(P));
        scanf("%d", &n);
        for (int i = 0 ; i < n; ++ i) {
            scanf("%d", &P[i].v);
            P[i].x = i + 1;
            add (P[i].x, P[i].v);
        }
        //cout << sum(3) << endl;
        printf("Case %d:
", ++ cur);
        while (scanf("%s", op)) {
            if (op[0] == 'E') {
                break;
            } else if (op[0] == 'Q') {
                int x, y;
                scanf("%d %d", &x, &y);
                cout << _sum(x, y) << endl;
            } else if (op[0] == 'S') {
                int x, y;
                scanf("%d %d", &x, &y);
                add (x, -y);
            } else if (op[0] == 'A') {
                int x, y;
                scanf("%d %d", &x, &y);
                add (x, y);
            }
        }
    }
    return 0;
}

#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[150050];

int n, num[50050] = {0};

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(2 * k, l, (l + r) / 2);
        build(2 * k + 1, ((l + r) / 2) + 1, r);
        node[k].sum = node[2 * k].sum + node[2 * k + 1].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[2 * k].r) {
        update(2 * k, i, x);
    }
    else update(2 * k + 1, 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[2 * k].r)  {
        return query(2 * k, l, r);
    }
    if (l >= node[2 * k + 1].l) {
        return query(2 * k + 1, l, r);
    } else {
        return query(2 * k, l, node[2 * k].r) + query(2 * k + 1, node[2 * k + 1].l, r);
    }
}

int main(){
    int T;
    scanf("%d",&T);
    for (int z = 1; z <= T; ++ z) {
        memset(num, 0, sizeof(num));
        scanf("%d", &n);
        for (int i = 1; i <= n; ++ i) {
            scanf("%d",&num[i]);
        }
        build(1, 1, n);
        char s[100];
        printf("Case %d:
",z);
        while (scanf("%s", s)){
            if (s[0] == 'E')    break;
            if (s[0] == 'Q'){
                int a, b;
                scanf("%d%d" ,&a ,&b);
                printf("%d
",query(1, a, b));
            }
            if (s[0] == 'A'){
                int a,x;
                scanf("%d%d", &a, &x);
                update(1, a, x);
            }
            if (s[0] == 'S') {
                int a,x;
                scanf("%d%d", &a, &x);
                update(1, a, -x);
            }
        }
    }
    return 0;
}