[JSOI2008]最大数 题解

前言

巨佬说:要有线段树，于是蒟蒻就开始打线段树。

蒟蒻只能拿之前0分的板子题开刀了QAQ。

题解

一开始我以为插入操作不带取模，于是打了下面这个弱智玩意
下面的代码是会WA的

#include <cstdio>
#include <algorithm>
#define ll long long

using namespace std;

ll read(){
	ll x = 0; int zf = 1; char ch = ' ';
	while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
	if (ch == '-') zf = -1, ch = getchar();
	while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); return x * zf;
}

ll s[800005];
ll _max[800005], cnt[800005];
int m, d, pos = 1;
const int MAXM = 200001;

pair<ll, ll> query(int pos, int l, int r, int x, int y){
	if (x <= l && r <= y)
		return make_pair(cnt[pos], _max[pos]);
	pair<ll, ll> ans = make_pair(-(1ll << 62), -(1ll << 62));
	int mid = (l + r) >> 1;
	if (x <= mid)
		ans = max(ans, query(pos << 1, l, mid, x, y));
	if (mid < y)
		ans = max(ans, query(pos << 1 | 1, mid + 1, r, x, y));
	return ans;
}

void add(int pos, int l, int r, int x, ll val){
	if (l == r){
		s[pos] += val, cnt[pos] += s[pos] / d; s[pos] %= d;
		_max[pos] = s[pos];
		return ;
	}
	int mid = (l + r) >> 1;
	if (x <= mid)
		add(pos << 1, l, mid, x, val);
	else if (mid < x)
		add(pos << 1 | 1, mid + 1, r, x, val);
	s[pos] = (s[pos << 1] + s[pos << 1 | 1]) % d;
	if (cnt[pos << 1] > cnt[pos << 1 | 1])
		_max[pos] = _max[pos << 1], cnt[pos] = cnt[pos << 1];
	else if (cnt[pos << 1] < cnt[pos << 1 | 1])
		_max[pos] = _max[pos << 1 | 1], cnt[pos] = cnt[pos << 1 | 1];
	else{
		cnt[pos] = cnt[pos << 1];
		_max[pos] = max(_max[pos << 1], _max[pos << 1 | 1]);
	}
}

int main(){
	m = read(), d = read(); char op[1]; ll t = 0;
	while (m--){
		scanf("%s", op); int n = read();
		if (op[0] == 'Q')
			printf("%lld
", t = query(1, 1, MAXM, pos - n + 1, pos).second);
		else if (op[0] == 'A')
			add(1, 1, MAXM, ++pos, n + t);
	}
	return 0;
}

一波上交WA 0。
然后一看不对啊，样例都过不了啊(我自信的没测样例)。
仔细看了一下题目，发现插入操作带取模。QAQ
简直有毒。
然后一遍过...

#include <cstdio>
#include <algorithm>
#define ll long long

using namespace std;

ll read(){
	ll x = 0; int zf = 1; char ch = ' ';
	while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
	if (ch == '-') zf = -1, ch = getchar();
	while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); return x * zf;
}

ll s[800005];
ll _max[800005];
int m, d, pos = 1;
const int MAXM = 200001;

ll query(int pos, int l, int r, int x, int y){
	if (x <= l && r <= y)
		return _max[pos];
	ll ans = -(1ll << 62);
	int mid = (l + r) >> 1;
	if (x <= mid)
		ans = max(ans, query(pos << 1, l, mid, x, y));
	if (mid < y)
		ans = max(ans, query(pos << 1 | 1, mid + 1, r, x, y));
	return ans;
}

void add(int pos, int l, int r, int x, ll val){
	if (l == r){
		(s[pos] += val) %= d, _max[pos] = s[pos];
		return ;
	}
	int mid = (l + r) >> 1;
	if (x <= mid)
		add(pos << 1, l, mid, x, val);
	else if (mid < x)
		add(pos << 1 | 1, mid + 1, r, x, val);
	s[pos] = (s[pos << 1] + s[pos << 1 | 1]) % d;
	_max[pos] = max(_max[pos << 1], _max[pos << 1 | 1]);
}

int main(){
	m = read(), d = read(); char op[1]; ll t = 0;
	while (m--){
		scanf("%s", op); int n = read();
		if (op[0] == 'Q')
			printf("%lld
", t = query(1, 1, MAXM, pos - n + 1, pos));
		else if (op[0] == 'A')
			add(1, 1, MAXM, ++pos, (n + t) % d);
	}
	return 0;
}