HDOJ 1166 敌兵布阵（线段树基本操作）

思路：

基本的线段树操作：单点更新，区间求和

#include <cstdio>
#include <cstdlib>
#include <cstring>

#define lhs l, m, rt << 1
#define rhs m + 1, r, rt << 1 | 1

const int maxn = 50010;
int seg[maxn << 2];

void PushUp(int rt)
{
    seg[rt] = seg[rt << 1] + seg[rt << 1 | 1];
}

void Build(int l, int r, int rt)
{
    if (l == r)
        scanf("%d", &seg[rt]);
    else
    {
        int m = (l + r) >> 1;
        Build(lhs);
        Build(rhs);
        PushUp(rt);
    }
}

void Update(int p, int delta, int l, int r, int rt)
{
    if (l == r)
        seg[rt] += delta;
    else
    {
        int m = (l + r) >> 1;
        if (p <= m)
            Update(p, delta, lhs);
        else
            Update(p, delta, rhs);
        PushUp(rt);
    }
}

int Query(int beg, int end, int l, int r, int rt)
{
    if (beg <= l && r <= end)
        return seg[rt];

    int m = (l + r) >> 1;
    int ret = 0;
    if (beg <= m)
        ret += Query(beg, end, lhs);
    if (end > m)
        ret += Query(beg, end, rhs);

    return ret;
}

int main()
{
    int cases;
    scanf("%d", &cases);

    for (int t = 1; t <= cases; ++t)
    {
        int n;
        scanf("%d", &n);
        Build(1, n, 1);

        printf("Case %d:\n", t);

        char op[16];
        while (scanf("%s", op) && op[0] != 'E')
        {
            int a, b;
            scanf("%d %d", &a, &b);

            if (op[0] == 'Q')
                printf("%d\n", Query(a, b, 1, n, 1));
            else if (op[0] == 'A')
                Update(a, b, 1, n, 1);
            else if (op[0] == 'S')
                Update(a, -b, 1, n, 1);
        }
    }
    return 0;
}