【BZOJ 1012】 [JSOI2008]最大数maxnumber(线段树做法)

【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
, 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;
}