2019CCPC秦皇島現場賽——[D-J-I]

問題鏈接來源與hduoj

問題蟲洞——D：Decimal

 

黑洞內窺：

輸入n，求1/n是否是無理數。若是輸出Yes，否則輸出No

(1<=n<=100)

 

思維光年：

由於n的值非常小，所以可以利用精度的損失來算

還有一種做法是包含2或者5的因子的數是有理數。。。不過，都可以。。

 

ACcode：

//#include<bits/stdc++.h>
#include  <stdio.h>
#include <iostream>
#include<algorithm>
#include      <map>
#include      <set>
#include   <vector>
#include    <queue>
#include    <stack>
#include <stdlib.h>
#include  <cstring>
#include <string.h>
#include   <string>
#include   <math.h>
using namespace std;
typedef long long ll;
#define MAXN 200
#define INF 0x3f3f3f3f//將近ll類型最大數的一半，而且乘2不會爆ll

int main()
{
    int t;
    cin >> t;
    while(t--)
    {
        int n;
        scanf("%d", &n);
        ll ans = 10000000000/n;        //這裏精度會損失，如果是無理數的話。。。
        if(ans*n == 10000000000) puts("No");
        else puts("Yes");
    }
    return 0;
}

 

問題蟲洞——J：MUV LUV EXTRA

 

黑洞內窺：

看這個鏈接吧，，，中文版的題目：MUV LUV EXTRA

求一個可靠值sum，可靠值：sum = a*(循環體在字符串裏出現的長度)-b*(循環體長度)。

 

思維光年：

將字符串倒過來，枚舉每個前綴串的i-next[i]即是循環體長度。

 

ACcode：

//#include<bits/stdc++.h>
#include  <stdio.h>
#include <iostream>
#include<algorithm>
#include      <map>
#include      <set>
#include   <vector>
#include    <queue>
#include    <stack>
#include <stdlib.h>
#include  <cstring>
#include <string.h>
#include   <string>
#include   <math.h>
using namespace std;
typedef long long ll;
#define MAXN 10000007
#define INF 0x3f3f3f3f//將近ll類型最大數的一半，而且乘2不會爆ll

int Next[MAXN];
int Next1[MAXN];
char ch[MAXN];
char sh[MAXN];
ll ans, a, b;
void NEXT(char *ch, int *Next)
{
    int lec = strlen(ch);
    int i=0, j=-1;
    Next[0] = -1;
    while(i < lec)
        if(j == -1 || ch[i] == ch[j]) 
            Next[++i] = ++j;
        else 
            j = Next[j];
}

int main()
{
    while(~scanf("%lld %lld", &a, &b))
    {
       memset(Next, 0, sizeof(Next));
       scanf("%s", sh);
       int len = strlen(sh);
       int x=0;
       for(int i=len-1; i>=0; --i)
           if(sh[i] == '.') break;
           else ch[x++] = sh[i];
       NEXT(ch, Next);
       ll sum = a-b;
       for(int i=1; i<=x; ++i)
       {
            ll cnt = a*(i) - b*(i-Next[i]);
            sum = max(sum, cnt);
       }
       cout << sum << '\n';
    }
    return 0;
}

 

問題蟲洞——I：Invoker

 

思維光年：

不太聰明的亞子：

枚舉六種狀態，貪心。。。

正確的求解：

dp

推薦博客：2019ccpc秦皇島 Invoker（dp / 遞推）

#include<stdio.h>
#include<iostream>
#include<map>
#include<algorithm>
#include<cstring>
#include<string.h>
#include<math.h>
using namespace std ;
typedef long long ll;
#define MAXN 100005
#define INF 0x3f3f3f3f//將近int類型最大數的一半，而且乘2不會爆int

int dp[MAXN][10];
char a[15][6][4]=
{
	{"QQQ","QQQ","QQQ","QQQ","QQQ","QQQ"},
	{"QQW","QWQ","WQQ","QQW","QQW","QQW"},
	{"QQE","QEQ","EQQ","QQE","QQE","QQE"},
	{"WWW","WWW","WWW","WWW","WWW","WWW"},
	{"QWW","WQW","WWQ","QWW","QWW","QWW"},
	{"WWE","WEW","EWW","WWE","WWE","WWE"},
	{"EEE","EEE","EEE","EEE","EEE","EEE"},
	{"QEE","EQE","EEQ","QEE","QEE","QEE"},
	{"WEE","EWE","EEW","WEE","WEE","WEE"},
	{"QWE","QEW","EQW","EWQ","WQE","WEQ"}
};

int is(string a, string b)
{
    if(a == b) return 0;
    else if(a[1] == b[0] && a[2] == b[1]) return 1; //增加的技能數
    else if(a[2] == b[0]) return 2;
    return 3;
}

int main ()
{
    ios::sync_with_stdio(false);
    string s;
    map<char, int>mp;
    mp['Y']=0;mp['V']=1;mp['G']=2;mp['C']=3;mp['X']=4;
    mp['Z']=5;mp['T']=6;mp['F']=7;mp['D']=8;mp['B']=9;
    while(cin >> s)
    {
        int n=s.size();
        s = " "+s;
        memset(dp, INF, sizeof(dp));
        for(int i=0; i<6; ++i)
            dp[1][i] = 3;
        for(int i=2; i<=n; ++i)
            for(int j=0; j<6; ++j)
            for(int k=0; k<6; ++k)
            dp[i][j] = min(dp[i][j], dp[i-1][k]+is(a[mp[s[i-1]]][k], a[mp[s[i]]][j]));
        int minn = INF;
        for(int i=0; i<6; ++i)
            minn = min(minn, dp[n][i]);
        cout << n+minn << '\n';
    }
    return 0;
}

 

10.21

秦皇島分站賽事已經過去很久了，可是仍然感覺到非常遺憾，

有點愧疚那天凌晨兩點時分的濤聲，不甘那四個半小時的無腦憨憨，

思來想去卻又無可奈何，躊躇滿志卻又不付諸行動.........