题意:
给你n个花瓶,一开始都没放花,现在有两个操作。
1.要求在从位置p开始往后放val个花,如果该花瓶有花,就不管这个花瓶往下看,若放完n个花瓶还有剩余,就丢掉,然后输出第一个放花的位置和最后一个放花的位置。
2.把a~b区间的花全都丢掉,然后输出丢掉花的数量。
思路:
先看操作1,就是通过前缀和思想找到第一个放花的位置和最后一个放花的位置, 这样好操作,不然找从p开始有第一个花的位置咋找吗。
操作二就用区间长度减去区间空瓶就是答案,然后更新一下就行。
ac代码:
/**
* Think twice, code once.
* 1.integer overflow(maybe even long long overflow : (a+b >= c) -> (a >= c-b)
* 2.runtime error
* 3.boundary condition
* ---------------------------------------------------------------------------------------
* Author : zzy
* Date : 2020-03-21-17.04.53 Saturday
*/
#include <bits/stdc++.h>
#define eb emplace_back
#define mp make_pair
#define mt make_tuple
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define forn(i, n) for (int i = 0; i < (int)(n); ++i)
#define for1(i, n) for (int i = 1; i <= (int)(n); ++i)
#define ford(i, a, b) for (int i = (int)(a); i >= (int)b; --i)
#define fore(i, a, b) for (int i = (int)(a); i <= (int)(b); ++i)
#define rep(i, l, r) for (int i = (l); i <= (r); i++)
#define per(i, r, l) for (int i = (r); i >= (l); i--)
#define ms(x, y) memset(x, y, sizeof(x))
#define SZ(x) ((int)(x).size())
using namespace std;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<pii> vpi;
typedef vector<vi> vvi;
typedef long long i64;
typedef vector<i64> vi64;
typedef vector<vi64> vvi64;
typedef pair<i64, i64> pi64;
typedef double ld;
template<class T> bool uin(T &a, T b) { return a > b ? (a = b, true) : false; }
template<class T> bool uax(T &a, T b) { return a < b ? (a = b, true) : false; }
const int maxn = 50001+1000;
int tc, n, m, no[maxn<<2], laz[maxn<<2];
void pushUp(int node) {
no[node] = no[node<<1]+no[node<<1|1];
}
void pushDown(int node, int l, int r) {
int mid = (l+r)>>1;
int lf = mid-l+1;
int rt = r-mid;
if (laz[node] == 1) {
no[node<<1] = no[node<<1|1] = 0;
laz[node<<1] = laz[node<<1|1] = laz[node];
laz[node] = -1;
}
if (laz[node] == 0) {
no[node<<1] = lf;
no[node<<1|1] = rt;
laz[node<<1] = laz[node<<1|1] = laz[node];
laz[node] = -1;
}
}
void build(int node, int l, int r) {
int len = r-l+1;
no[node] = len;
laz[node] = -1;
if (l < r) {
int mid = (l+r)>>1;
build(node<<1, l, mid);
build(node<<1|1, mid+1, r);
}
}
int querySum(int node, int l, int r, int ql, int qr) {
if (l >= ql && r <= qr) return no[node];
pushDown(node, l, r);
int sum = 0;
int mid = (l+r)>>1;
if (ql <= mid) sum += querySum(node<<1, l, mid, ql, qr);
if (qr > mid) sum += querySum(node<<1|1, mid+1, r, ql, qr);
return sum;
}
int queryPos(int node, int l, int r, int val) {
if (l == r) return l;
pushDown(node, l, r);
int mid = (l+r)>>1;
if (no[node<<1] >= val) queryPos(node<<1, l, mid, val);
else queryPos(node<<1|1, mid+1, r, val-no[node<<1]);
}
void updateOne(int node, int l, int r, int ul, int ur) {
if (l >= ul && r <= ur) {
no[node] = 0;
laz[node] = 1;
return;
}
pushDown(node, l, r);
int mid = (l+r)>>1;
if (ul <= mid) updateOne(node<<1, l, mid, ul, ur);
if (ur > mid) updateOne(node<<1|1, mid+1, r, ul, ur);
pushUp(node);
}
void updateZero(int node, int l, int r, int ul, int ur) {
if (l >= ul && r <= ur) {
no[node] = r-l+1;
laz[node] = 0;
return;
}
pushDown(node, l, r);
int mid = (l+r)>>1;
if (ul <= mid) updateZero(node<<1, l, mid, ul, ur);
if (ur > mid) updateZero(node<<1|1, mid+1, r, ul, ur);
pushUp(node);
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
cout.precision(10);
cout << fixed;
#ifdef LOCAL_DEFINE
freopen("input.txt", "r", stdin);
#endif
cin >> tc;
while (tc--) {
cin >> n >> m;
build(1, 1, n);
forn(i, m) {
int op, x, y, pos, val;
cin >> op;
if (op == 1) {
cin >> pos >> val;
++pos;
int res = querySum(1, 1, n, pos, n);
if (!res) cout << "Can not put any one." << endl;
else {
int pref = pos==1?0:querySum(1, 1, n, 1, pos-1);
int sum;
if (res < val) sum = res;
else sum = val;
int fp = queryPos(1, 1, n, pref+1);
int sp = queryPos(1, 1, n, pref+sum);
cout << fp-1 << ' ' << sp-1 << '\n';
updateOne(1, 1, n, fp, sp);
}
} else {
cin >> x >> y;
++x;
++y;
cout << (y-x+1-querySum(1, 1, n, x, y)) << endl;
updateZero(1, 1, n, x, y);
}
}
cout << '\n';
}
#ifdef LOCAL_DEFINE
cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
#endif
return 0;
}