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;
}