正睿2018暑期集训CD班附加赛Round4暨贺爸探底纪念赛123民间题解
BY QYJ

T1:不知道什么鬼~
std:

#include<bits/stdc++.h>
using namespace std;
int t;
double x;
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lf",&x);
        printf("%.10lf\n",(double)1/(1-x));
    }
    return 0;
}

T2:大模拟
将数字分为4段（歪果仁都是三位一段的）
每一段内进行操作。
预处理出1~9，10~19，10，20，30，40，50，60，70，80，90，100
即可操作
std:

#include<bits/stdc++.h>
using namespace std;
int t; 
string s[11]={"zero","one","two","three","four","five","six","seven","eight","nine"};
string p[11]={"ten","twenty","thirty","forty","fifty","sixty","seventy","eighty","ninety"};
string Q="hundred";
string l[11]={"ten","eleven","twelve","thirteen","fourteen","fifteen","sixteen","seventeen","eighteen","nineteen"};
void put(string s)
{
    for(int i=0;i<s.size();i++)
        printf("%c",s[i]);
    return;
}
void print(int x)
{
    if(x==0) return;
    else if(x<10) 
    {
        put(s[x]);
        printf(" ");
        return;
    }
    else if(x<20)
    {
        put(l[x-10]);
        printf(" ");
        return;
    }
    else if(x<100)
    {
        put(p[x/10-1]);
        if(x%10!=0)
        {
            printf("-");
            put(s[x%10]);
        }
        printf(" ");
        return;
    }
    else if(x%100==0)
    {
        put(s[x/100]);
        printf(" ");
        put(Q);
        printf(" ");
        return;
    }
    else
    {
        put(s[x/100]);
        printf(" ");
        put(Q);
        printf(" and ");
        if((x/10)%10-1>0) 
        {
            put(p[(x/10)%10-1]);
            if(x%10!=0) 
            {
                printf("-");
            }
        }
        if(x%10!=0) 
        {
            put(s[x%10]);
        }
        printf(" ");
        return;
    }
}
int main()
{
    //外文数字：3位一断 
    //x and y-z xxx
    //例如 one hundred and twenty-three million 
    //分3个档做：
    //million
    //thousand 
    //one 
    //打表暴力从0~999
    //freopen("大样例-送温暖.out","w",stdout);
    scanf("%d",&t);
    while(t--)
    {
        int x;
        scanf("%d",&x);
        if(x==0)
        {
            printf("zero\n");
            continue;
        }
        if(x>999999999) 
        {
            put(s[x/1000000000]);
            printf(" billion ");
        }
        if((x/1000000)%1000!=0)
        {
            print((x/1000000)%1000);
            if(x>=1000000) printf("million ");
        } 
        if((x/1000)%1000!=0)
        {
            print((x/1000)%1000);
            if(x>=1000) printf("thousand ");
        }
        print(x%1000); 
        printf("\n");
    }
    return 0;
}

T3:
13pts:
这题明显可以dp。
设f[i]表示在第i个点时最小消耗了几块钱。
dfs转移。
33~100pts:
将它看成有向图。
前后两个点连一条边。
u->v的边权即为v
haha
记得加上起始位置的点权！
你寇以用SPFA加上玄学优化
当然......
100pts:
dijstra+堆优化 才是正义的！