NOIP模拟六

合集 - 模拟赛(23)

1.12.30模拟赛2023-12-302.12.23模拟赛2023-12-233.国庆模拟赛一总结2023-10-014.国庆模拟赛二总结2023-10-025.国庆模拟赛三总结2023-10-036.国庆模拟赛四2023-10-057.NOIP模拟一2023-10-098.NOIP模拟二2023-10-109.NOIP模拟三2023-10-1110.NOIP模拟四2023-10-1211.NOIP模拟赛五2023-10-16

12.NOIP模拟六2023-10-17

13.NOIP模拟七2023-10-1814.NOIP模拟八2023-10-1915.20240223【省选】模拟2024-02-2316.20240420【提高】模拟2024-04-2017.20240724【省选】模拟2024-07-2418.20240621【省选】模拟2024-06-2119.20240718【提高】模拟2024-07-1820.20240726【省选】模拟2024-07-2621.20240917【省选】模拟2024-10-2122.20241112【NOIP】模拟2024-11-1223.20241118【NOIP】模拟2024-11-18

收起

又是大寄的一天，真就只有提高二等水平。

link

T1

分类讨论，过了

code

点击查看代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const ll N=114514,M=1919810;
ll T;
char s[N],t[N];
ll val[N];
int main(){
	//freopen("T1.in","r",stdin);
	//freopen("T1.out","w",stdout);
	cin>>T;
	while(T--){
		cin>>(s+1);
		ll n=strlen(s+1),len=strlen(s+1);
		ll ans=4;
		for(int i=1;i<=n;++i){
			if(s[i]=='n'||s[i]=='o'||s[i]=='i'||s[i]=='p') ans=min(ans,3ll);
			if((s[i]=='n'&&s[i+1]=='o')||(s[i]=='n'&&s[i+1]=='i')||(s[i]=='n'&&s[i+1]=='p')
			||(s[i]=='o'&&s[i+1]=='i')||(s[i]=='o'&&s[i+1]=='p')||(s[i]=='i'&&s[i+1]=='p')) ans=min(ans,2ll);
			if((s[i]=='n'&&s[i+1]=='o'&&s[i+2]=='i')||(s[i]=='n'&&s[i+1]=='o'&&s[i+2]=='p')
			||(s[i]=='n'&&s[i+1]=='i'&&s[i+2]=='p')||(s[i]=='o'&&s[i+1]=='i'&&s[i+2]=='p')) ans=min(ans,1ll);
			if(s[i]=='n'&&s[i+1]=='o'&&s[i+2]=='i'&&s[i+3]=='p') ans=min(ans,0ll);
			if(s[i]=='o'&&s[i+1]=='n'&&s[i+2]=='i'&&s[i+3]=='p') ans=min(ans,1ll);
			if(s[i]=='n'&&s[i+1]=='i'&&s[i+2]=='o'&&s[i+3]=='p') ans=min(ans,1ll);
			if(s[i]=='n'&&s[i+1]=='o'&&s[i+2]=='p'&&s[i+3]=='i') ans=min(ans,1ll);
			if(s[i]=='o'&&s[i+1]=='n'&&s[i+2]=='p'&&s[i+3]=='i') ans=min(ans,2ll);
			if(s[i-1]=='n'&&s[i+2]=='p') ans=min(ans,2ll);
			if(s[i-1]=='n'&&s[i+1]=='o') ans=min(ans,3ll);
			if(s[i-1]=='o'&&s[i+1]=='i') ans=min(ans,3ll);
			if(s[i-1]=='i'&&s[i+1]=='p') ans=min(ans,3ll);
			if(s[i-1]=='n'&&s[i+1]=='i') ans=min(ans,2ll);
			if(s[i-1]=='o'&&s[i+1]=='p') ans=min(ans,2ll);
			if(s[i-1]=='n'&&s[i+1]=='p') ans=min(ans,2ll);
			if(s[i-2]=='n'&&s[i-1]=='o'&&s[i+1]=='i'&&s[i+2]=='p') ans=min(ans,1ll);
			if(s[i-2]=='n'&&s[i-1]=='o'&&s[i+2]=='i'&&s[i+3]=='p') ans=min(ans,2ll);
			if(s[i-1]=='n'&&s[i+1]=='i'&&s[i+2]=='p') ans=min(ans,1ll);
			if(s[i-2]=='n'&&s[i-1]=='o'&&s[i+1]=='p') ans=min(ans,1ll);
		}
		cout<<ans<<'\n';
	}
	return 0;
}

T2

想到用数据结构，但是不知道维护啥……

思路：先找到每一行每一列的编号最小的障碍物，各自对应列、行的编号，然后先把能加上的格子都给加上，但是这样会有重复的，要考虑怎么去重（我就是这一步不会了）。

解法：用一个数据结构快速维护前缀和与查询，但是原理是什么啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊

妈的不搞了之后问同学……

T3&T4

T3分组dp容斥系数预处理，T4不知道，都不会。