【题目链接】:http://www.lydsy.com/JudgeOnline/problem.php?id=1012

【题意】

【题解】



预开一个20W长度的线段树;

这里a[1..20W]={0};

每次对单个点进行修改操作;

然后返回区间的最大值;



【完整代码】

/**************************************************************

    Problem: 1012

    User: chengchunyang

    Language: C++

    Result: Accepted

    Time:1032 ms

    Memory:9104 kb

****************************************************************/

#include <bits/stdc++.h>

using namespace std;

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define LL long long

#define rep1(i,a,b) for (int i = a;i <= b;i++)

#define rep2(i,a,b) for (int i = a;i >= b;i--)

#define mp make_pair

#define pb push_back

#define fi first

#define se second

#define rei(x) scanf("%d",&x)

#define rel(x) scanf("%lld",&x)

typedef pair<int, int> pii;

typedef pair<LL, LL> pll;

const int dx[9] = { 0,1,-1,0,0,-1,-1,1,1 };

const int dy[9] = { 0,0,0,-1,1,-1,1,-1,1 };

const double pi = acos(-1.0);

const int N = 2e5+100;

LL a[N], ma[N << 2],d,lastans = 0,ne_w;

int m,len = 0;

void up_data(int pos, int l, int r, int rt)

{

    if (l == r)

    {

        ma[rt] = ne_w;

        return;

    }

    int m = (l + r) >> 1;

    if (pos <= m)

        up_data(pos, lson);

    else

        up_data(pos, rson);

    ma[rt] = max(ma[rt << 1], ma[rt << 1 | 1]);

}

LL query(int L, int R, int l, int r, int rt)

{

    if (L <= l && r <= R)

        return ma[rt];

    int m = (l + r) >> 1;

    LL temp1 = -1, temp2 = -1;

    if (L <= m)

        temp1 = query(L, R, lson);

    if (m < R)

        temp2 = query(L, R, rson);

    return max(temp1, temp2);

}

int main()

{

    //freopen("F:\\rush.txt", "r", stdin);

    rei(m), rel(d);

    rep1(i, 1, m)

    {

        char key; LL t;

        key = getchar();

        scanf("%c %lld", &key, &t);

        if (key == 'A')

        {

            ne_w = (t + lastans) % d;

            len++;

            up_data(len, 1, 200000, 1);

        }

        else

            lastans = query(len - t + 1, len, 1, 200000, 1), printf("%lld\n", lastans);

    }

    return 0;

}