hdu 4288(区间合并)

题意：给出三种操作，add x表示往序列里添加x，del x表示将序列中的x删除，sum要求输出序列按升序排列好后中下标%5==3的数字的和。
题解：线段树每个节点要存两种值，一个是当前区间内还有多少个数字，另一个是sum[i]表示下标%5==i的数字的和，修改数字比较很简单，而在同时维护sum[i]的值比较不好办，一个节点的sum[i]可以由两个孩子节点得到，tree[k].sum[(i+tree[k*2].num)%5] = tree[k*2].sum[(i+tree[k*2].num)%5] + tree[k*2+1].sum[i]，因为将右边的下标确定，左边的可能会一些没有下标的数字被删掉了，应该按数量+i计算。

#include 
#include 
#include 
using namespace std;
const int N = 100005;
struct Tree {
    int num;
    long long sum[5];
}tree[N << 2];
int a[N], b[N], n;
char str[N][10];

void pushup(int k) {
    for (int i = 0; i < 5; i++) {
        int temp = (i + tree[k * 2].num) % 5;
        tree[k].sum[temp] = tree[k * 2].sum[temp] + tree[k * 2 + 1].sum[i];
    }
}

void build(int k, int left, int right) {
    tree[k].num = 0;
    memset(tree[k].sum, 0, sizeof(tree[k].sum));
    if (left == right)
        return;
    int mid = (left + right) / 2;
    build(k * 2, left, mid);
    build(k * 2 + 1, mid + 1, right);

}

void modify(int k, int left, int right, int pos, int x, int flag) {
    if (flag)
        tree[k].num++;
    else
        tree[k].num--;
    if (left == right) {
        tree[k].sum[1] = flag * x;  
        return;
    }
    int mid = (left + right) / 2;
    if (pos <= mid)
        modify(k * 2, left, mid, pos, x, flag);
    else
        modify(k * 2 + 1, mid + 1, right, pos, x, flag);
    pushup(k);
}

int main() {
    while (scanf("%d", &n) == 1) {
        int cnt = 0;
        for (int i = 1; i <= n; i++) {
            scanf("%s", str[i]);
            if (str[i][0] != 's') {
                scanf("%d", &b[i]);
                a[cnt++] = b[i];
            }
        }
        sort(a, a + cnt);
        cnt = unique(a, a + cnt) - a;
        if (cnt == 0)
            memset(tree[1].sum, 0, sizeof(tree[1].sum));
        else
            build(1, 1, cnt);
        for (int i = 1; i <= n; i++) {
            if (str[i][0] == 'a') {
                int pos = lower_bound(a, a + cnt, b[i]) - a + 1;
                modify(1, 1, cnt, pos, b[i], 1);
            }
            else if (str[i][0] == 'd') {
                int pos = lower_bound(a, a + cnt, b[i]) - a + 1;
                modify(1, 1, cnt, pos, b[i], 0);
            }
            else
                printf("%lld\n", tree[1].sum[3]);
        }
    }
    return 0;
}