题目

A. Mark the Photographer

题意:

思路:

code:

B. Mark the Dust Sweeper

题意:

思路:

code:

C. Mark and His Unfinished Essay

题意:

思路:

code:

总结:

题目

A. Mark the Photographer

题意:

给你 2*n个数, 一个数x, 要求把这2n个数分成2组, 保证其中一组每个数都比另一组的数大x, 即 $b_i-a_i>=x$ .

思路:

把这2n个数sort下, 截取成两段, 逐个对应判断是否存在 $b_i-a_i>=x$ 即可, 有点贪心思想.

code:

/***    author:  CurleyD*    created: 07.15.2022 21:35:34
**/
#include <bits/stdc++.h>
#define endl '\n'
#define IO ios::sync_with_stdio(false); cin.tie(nullptr);
using namespace std;
typedef long long LL;
//#define LOCAL
//#define int LL
//headconst int INF = 0x3f3f3f3f;
const int N = 1e5 + 10;
int T, n, m, x;int a[N];void solve() {cin >> n >> x;for (int i = 1; i <= n * 2; i++) cin >> a[i];sort(a+1,a+1+2*n);for (int i = 1; i <= n; i++) {if (a[i+n] - x >= a[i]) {continue;}else {cout << "NO\n";return;}}cout<<"YES\n";
}signed main() {IO;#ifdef LOCALfreopen("out.txt","w",stdout);#endifcin >> T;while(T--) {solve();}return 0;
}

B. Mark the Dust Sweeper

题意:

题目给的题意不明确, 大概意思就是给你个n个正数, 对于 1~n 你可以任选 i, j在这范围之内

进行 ai = ai - 1, aj = aj + 1, (需要保证为正), 问你最少多少次使得 1~n-1的数全为0? 描述的可能不太清晰...

思路:

根据下面的提示来看似乎不能一直选j=n???

观察样例 + 猜想 + 结合实际吸尘器不能跳着吸只能顺着推可知, ans = (1~n-1的ai) + (1~n-1相邻的正数相差的0的数量即间隔数);

不知道怎么证明???

code:

/***    author:  CurleyD*    created: 07.15.2022 21:57:42
**/
#include <bits/stdc++.h>
#define endl '\n'
#define IO ios::sync_with_stdio(false); cin.tie(nullptr);
using namespace std;
typedef long long LL;
//#define LOCAL
//#define int LL
//headconst int INF = 0x3f3f3f3f;
const int N = 2e5 + 10;
int T, n, m;int a[N];void solve() {cin >> n;LL res = 0;LL cnt = 0;for (int i = 1; i <= n; i++)cin >> a[i];bool f = 0;for (int i = 1; i <= n - 1; i++) {if (a[i] != 0) {f = 1;res = res + a[i] + cnt;cnt = 0;}if (a[i] == 0 && f) {cnt ++;}}    res += cnt;cout << res << endl;
}signed main() {IO;#ifdef LOCALfreopen("out.txt","w",stdout);#endifcin >> T;while(T--) {solve();}return 0;
}

C. Mark and His Unfinished Essay

题意:

给你个长度n的字符串, 每次可截取l, r 放在字符串的末尾使得字符串的长度跟着增长, 进行完上述操作给你k个询问, 每个询问给你个位置, 输出增长完的字符串的对应字符是什么?

思路:

l, r 都是1e18显然模拟会超时, 开始一直想着hash来做, 但是我不会

我的做法可能会fst...但pretest过了

通过他最大每次可能给30次修改, 想到可以溯源每个pos映射在原来s串的位置

具体解释下的话如下:

一个样例:n=4,m=3,k=1,s="mark"

4 3 1
mark
1 4
5 7
3 8
10

三次修改, 一次查询

我们用newl表示新增区间l, newr同理, len表示新增区间的长度.

markmark l = 1, r = 4, newl = 5, newr = 8, len = 4

markmarkmar l = 5, r = 7, newl = 9, newr = 11, len = 3

markmarkrkmark l = 3, r = 8, newl = 12, newr = 17, len = 6

对于pos = 10, 它是由l=5,r=7区间变成的, 则pos相当于其中的 x = 6!

因此我们可以把pos设置成pos = 6!!!

同理pos=6 -> pos=newl - l + pos!!!

因此我们的过程大致为:

1)读入s,n,m,k, 设置SZ=s.size() (以后要更新会增长)

2)读入l, r 对于每组l,r, 记录其新老l,r到结构体数组p中即len, 更新SZ

3)sort下p, 方便之后找, 当然不sort估计也行

4)查询过程, 对于每个pos, 进行溯源, 直到pos 在 1~n之间为止!

得解

要是没有fst那证明这个做法可行的, 但感觉应该是对的.

code:

/***    author:  CurleyD*    created: 07.15.2022 22:08:18
**/
#include <bits/stdc++.h>
#define endl '\n'
#define IO ios::sync_with_stdio(false); cin.tie(nullptr);
using namespace std;
typedef long long LL;
//#define LOCAL
//#define int LL
//headconst int INF = 0x3f3f3f3f;
const int N = 1e5 + 10;
int T, n, m, k;LL l, r, pos;struct node{LL l, r, len;LL x, y;
}p[55];bool cmp(node a, node b) {return a.l < b.l;
}void solve() {string s;cin >> n >> m >> k;cin >> s;LL SZ = s.size();s.insert(s.begin(),'#');for (int i = 1; i <= m; i++) {cin >> l >> r;LL len = r - l + 1;//老区间p[i].x = l;p[i].y = r;//新区间p[i].l = SZ + 1;p[i].r = SZ + len;p[i].len = len;SZ += len;//cout<<p[i].l << " " << p[i].r << endl;}sort(p + 1, p + 1 + m, cmp);for (int q = 1; q <= k; q++) {cin >> pos;while (pos > n) {for (int i = 1; i <= m; i++) {if (p[i].l <= pos && p[i].r >= pos) {pos = p[i].x + pos - p[i].l;// cout << pos << endl;break;}}}cout<<s[pos]<<endl;}
}signed main() {IO;#ifdef LOCALfreopen("out.txt","w",stdout);#endifcin >> T;while(T--) {solve();}return 0;
}/*
5->7
9->11  10 - 9 + 5 -> 6
*/

总结:

1)简单题思路形成慢,

2)过于依赖翻译

3)有时候写一半发现漏掉条件/读了假题 ->这次的B

4)会在细节出问题, ->这次的C, 结构体里忘记开LL, len忘开LL导致出现负数溢出, 血亏.

毕竟是菜狗, 第一次赛时出c已经挺高兴了, D似乎是个dp? 看了下tourist的不是dp...

update:

开测了QAQ

没FST, 好耶LOL

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Codeforces Round #807 (Div. 2) A-C

题目

A. Mark the Photographer

题意:

思路:

code:

B. Mark the Dust Sweeper

题意:

思路:

code:

C. Mark and His Unfinished Essay

题意:

思路:

code:

总结:

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