HDU4288 Coder 线段树

通过离线处理，由于线段树不能够动态的扩张，将所有的数都进行永久标号，无视信息的冗余。对于每一个节点，保留对5取余的所有余数的和值，用long long存储。然后根据元素个数进行更新。

代码如下：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<set>
#include<map>
#include<cstring>
#include<vector>
#include<string>
#define MAXN 100005
using namespace std;

typedef long long Int64;

char op[MAXN][5];
int num[MAXN], rec[MAXN], idx;

map<int,int>mp;

struct Node {
    int l, r, tot;
    Int64 a[5];
}s[MAXN*4];

void build(int p, int l, int r) {
    s[p].l = l, s[p].r = r,    s[p].tot = 0;
    memset(s[p].a, 0, sizeof (s[p].a));
    if (l != r) {
        int mid = (l + r) >> 1;
        build(p<<1, l, mid);
        build(p<<1|1, mid+1, r);
    }
}

void push_up(int p) {
    s[p].tot = s[p<<1].tot + s[p<<1|1].tot;
    for (int i = 0; i < 5; ++i) {
        int k = (s[p<<1].tot+i) % 5;
        s[p].a[k] = s[p<<1].a[k] + s[p<<1|1].a[i];
    }
}

void modify(int p, int pos, int f) {
    if (s[p].l == s[p].r) {
        if (f > 0) { 
            s[p].tot = 1, s[p].a[1] = rec[pos-1];
            // rec[pos-1]保留的是原来的值
        } else {
            s[p].tot = 0; s[p].a[1] = 0;
        }
    } else {
        int mid = (s[p].l + s[p].r) >> 1;
        if (pos <= mid) {
            modify(p<<1, pos, f);
        } else {
            modify(p<<1|1, pos, f);
        }
        push_up(p);
    }
}

int main()
{
    int N;
    while (scanf("%d", &N) != EOF) {
        idx = -1;
        mp.clear();
        for (int i = 1; i <= N; ++i) {
            scanf("%s", op[i]);
            if (op[i][0] == 'a') {
                scanf("%d", &num[i]);
                rec[++idx] = num[i];
            } else if (op[i][0] == 'd') {
                scanf("%d", &num[i]);
                rec[++idx] = num[i];
            }
        }
        sort(rec, rec+idx+1);
        idx = unique(rec, rec+idx+1) - rec; // 返回的是元素个数 
        for (int i = 0; i < idx; ++i) {
            mp[rec[i]] = i + 1; // 对数字进行离散化  
        }
        if (idx != -1) {
            build(1, 1, idx);
        }
        for (int i = 1; i <= N; ++i) {
            if (op[i][0] == 'a') {
                modify(1, mp[num[i]], 1);
            } else if (op[i][0] == 'd'){
                modify(1, mp[num[i]], -1);
            } else {
                printf("%I64d\n", s[1].a[3]);
            }
        }
    }
    return 0;
}