GCJ2019 Qualification Round abc题解

A.Foregone Solution
因为肯定有4，只要把4那位减1出来就行了。。。抱歉我是个傻子，退役太久，智商为负，写了个又麻烦又低效的东西，不想改了贴在这里纪念下sb时刻

#include<bits/stdc++.h>
#define mpi(i,j) make_pair(i,j)
using namespace std;
typedef pair<int,int> pll;
char str[105];
pll sol[10][2];
int ans[105][2][2];
int s[105][2];
int main(){
    int t;
    scanf("%d",&t);
    for(int i=0;i<10;i++){
        sol[i][0]=sol[i][1]=mpi(-1,-1);
    }
    for(int i=0;i<10;i++){
        for(int j=0;j<10;j++){
            if(i==4||j==4)continue;
            int p=i+j;
            if(p>=10)sol[p%10][1]=mpi(i,j);
            else sol[p][0]=mpi(i,j);
        }
    }
    for(int cas=1;cas<=t;cas++){
        scanf("%s",str);
        int ls=strlen(str);
        int cnt=0;
        for(int i=ls-1;i>=0;i--){
            int p=str[i]-'0';
            if(i!=ls-1){
                if((ans[i+1][1][0]||ans[i+1][1][1])){
                    ans[i][0][1]=sol[(p+9)%10][0].first==-1?0:1;
                    ans[i][1][1]=sol[(p+9)%10][1].first==-1?0:1;
                }
                if((ans[i+1][0][0]||ans[i+1][0][1])){
                    ans[i][0][0]=sol[p][0].first==-1?0:1;
                    ans[i][1][0]=sol[p][1].first==-1?0:1;
                }
            }else{
                ans[i][0][0]=sol[p][0].first==-1?0:1;
                ans[i][1][0]=sol[p][1].first==-1?0:1;
            }
        }
        int up=0;
        for(int i=0;i<ls;i++){
            int p=str[i]-'0';
            for(int j=0;j<((i==ls-1)?1:2);j++){
                if(ans[i][up][j]){
                    s[i][0]=sol[(p-j+10)%10][up].first;
                    s[i][1]=sol[(p-j+10)%10][up].second;
                    up=j;break;
                }
            }
        }
        printf("Case #%d: ",cas);
        int st;
        st=0;while(!s[st][0])st++;
        for(int i=st;i<ls;i++)printf("%d",s[i][0]);printf(" ");
        st=0;while(!s[st][1])st++;
        for(int i=st;i<ls;i++)printf("%d",s[i][1]);printf("\n");
    }
    return 0;
}

B.You Can Go Your Own Way
只需要E<=>S

#include<bits/stdc++.h>
using namespace std;
char str[100005];
int main(){
    int t;
    scanf("%d",&t);
    for(int cas=1;cas<=t;cas++){
        int n;
        printf("Case #%d: ",cas);
        scanf("%d%s",&n,str);
        for(int i=0;i<2*n-2;i++){
            if(str[i]=='S')printf("E");
            else printf("S");
        }puts("");
    }
    return 0;
}

C.Cryptopangrams
假设 $a\ b\ c$ 均为质数且
$a \neq c$ 则 $gcd(a*b,b*c)=b$

import java.math.BigInteger;
import java.util.*;
import java.io.*;
 
public class Solution
{
	public static void main(String args[])
	{
		Scanner in = new Scanner(System.in);
		BigInteger mul[] = new BigInteger[105];
		BigInteger ans[] = new BigInteger[105];
		BigInteger ans2[] = new BigInteger[105];
		Map<BigInteger, Character> mic = new HashMap<BigInteger, Character>();
		int t = in.nextInt();		
		for(int cas = 1; cas <= t; ++cas)
		{
		    System.out.print("Case #"+cas+": ");
		    BigInteger n=in.nextBigInteger();
		    int l=in.nextInt(),split=2,cnt=0;mic.clear();
		    for(int i=1;i<=l;i++)mul[i]=in.nextBigInteger();
		    while(split<=l){if(!mul[split].equals(mul[split-1]))break;split++;}
            ans2[split]=ans[split]=mul[split].gcd(mul[split-1]);
            for(int i=split-1;i>=1;i--)ans2[i]=ans[i]=mul[i].divide(ans[i+1]);
            for(int i=split+1;i<=l+1;i++)ans2[i]=ans[i]=mul[i-1].divide(ans[i-1]);
            Arrays.sort(ans2,1,l+2);
            for(int i=1;i<=l+1;i++){
                if(mic.containsKey(ans2[i]))continue;
                mic.put(ans2[i],(char)('A'+cnt++));
            }
            for(int i=1;i<=l+1;i++)System.out.print(mic.get(ans[i]));
            System.out.println("");
		}
	}
}

需要用到大数，学习了一下java

BigInteger官方文档

Array.sort(object,st,ed);(左闭右开)
Map<BigInteger,Charactor>xx = new HashMap<BigInteger,Charactor>
xx.containsKey/.get/.put
强制转换

GCJ2019 Qualification Round abc题解

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