APAC Practice

2016 round A 

A. Googol String

small && large

LL a[65];
int dfs(LL pos, int id, bool f)
{
    if(pos==1 || pos==a[id-1]+1)
        return f? 0:1;
    if(pos>a[id-1])
        return dfs(a[id]-pos+1, id-1, f^1);
    else
        return dfs(pos, id-1, f);
}
int main()
{
    freopen("A-small-practice.in", "r", stdin);
    freopen("A.out", "w", stdout);
    a[1]=1;
    for(int i=2;i<=60;i++)
        a[i]=a[i-1]*2+1;
    int t, ca=1;
    scanf("%d", &t);
    while(t--)
    {
        LL n;
        scanf("%I64d", &n);
        int pos=lower_bound(a, a+61, n)-a;
        printf("Case #%d: %d\n", ca++, dfs(n, pos, 1));
    }
    return 0;
}

 

B. gCube

 

n个数，m个询问

 

每个询问给[l, r]

 

求 $\sqrt[^{r-l+1}]{\prod\limits_{i=l}^{r}a_i}$

 

smal

 

暴力二分，large TLE了 跑10min都没跑完，应该是大数乘太费时间

 

public class Main
{
    static BigDecimal quick(BigDecimal a, BigInteger b)
    {
        BigDecimal ans=BigDecimal.ONE;
        while(b.compareTo(BigInteger.ZERO)!=0)
        {
            if(b.mod(BigInteger.valueOf(2)).compareTo(BigInteger.ZERO)!=0)
                ans=ans.multiply(a);
            b=b.divide(BigInteger.valueOf(2));
            a=a.multiply(a);
        }
        return ans;
    }
    static BigDecimal sqrt(BigDecimal x, int n)
    {
        BigDecimal l=BigDecimal.ZERO, r=x;
        while(r.subtract(l).abs().compareTo(BigDecimal.valueOf(1e-10))>0)
        {
            BigDecimal m=(l.add(r)).divide(BigDecimal.valueOf(2));
            BigDecimal cur=quick(m, BigInteger.valueOf(n));
            cur=cur.subtract(x);
            if(cur.abs().compareTo(BigDecimal.valueOf(1e-10))<0)
                return m;
            else if(cur.compareTo(BigDecimal.ZERO)<0)
                l=m;
            else 
                r=m;
        }
        return l;
    }
    public static void main(String[] args) throws FileNotFoundException 
    {
//        InputReader in=new InputReader();
//        PrintWriter out=new PrintWriter(System.out);
        Scanner in=new Scanner(new File("B-large-practice.in"));
        PrintWriter out = new PrintWriter(new File("B.out"));
        int t=in.nextInt();
        int ca=1;
        while((t--)!=0)
        {
            int n=in.nextInt();
            int m=in.nextInt();
            int []a=new int[n];
            for(int i=0;i<n;i++)
                a[i]=in.nextInt();
            BigDecimal []b=new BigDecimal[n];
            b[0]=BigDecimal.valueOf(a[0]);
            for(int i=1;i<n;i++)
                b[i]=b[i-1].multiply(BigDecimal.valueOf(a[i]));
            out.println("Case #"+ca+":");ca++;
            while((m--)!=0)
            {
                int l=in.nextInt();
                int r=in.nextInt();
                BigDecimal pre=BigDecimal.ONE;
                if(l!=0) pre=b[l-1];
                BigDecimal s=b[r].divide(pre);
                out.printf("%.9f", sqrt(s, r-l+1));
                out.println("");
            }
        }
        out.close();
    }
}

 

large

 

求n次根号下的就是把ans除n次刚好除完

 

那就for l 到 r 每次都除一个ans 最后与1比较 判断是否除完就行了

 

const LD eps=1e-10; // eps<1e9了要long double！！！

int a[1005];
LD sqrtn(int x, int y)
{
    LD l=1, r=a[y];
    for(int i=x;i<=y;i++)     // 数列不递增！！注意取最大值！
        r=max(r, (LD)a[i]);
    while(fabs(l-r)>eps)
    {
        LD mid=(l+r)/2.0;
        LD cur=1.0;
        for(int i=x;i<=y;i++)
            cur*=a[i]/mid;
　　　　// 要注意这里不能fabs(cur-1)<eps了就结束 因为cur与1<eps了 不代表mid与ans的误差也小于eps了！！
        if(cur>1)
            l=mid;
        else
            r=mid;
    }
    return l;
}
int main()
{
    freopen("B-large-practice.in", "r", stdin);
    freopen("B.out", "w", stdout);
    int t, ca=1;
    scanf("%d", &t);
    while(t--)
    {
        int n, m;
        scanf("%d%d", &n, &m);
        for(int i=0;i<n;i++)
            scanf("%d", &a[i]);
        printf("Case #%d:\n", ca++);
        while(m--)
        {
            int l, r;
            scanf("%d%d", &l, &r);
            printf("%.9Lf\n", sqrtn(l, r));
        }
    }
    return 0;
}

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2017 round A

A. Country Leader

给n个字符串 输出其中字母最多的字符串 字母一样多 输出字典序最小

神tm 空格only appear in Large！！ 挂了A large真是日了狗的心情

 1 char s[25];
 2 bool vis[1005];
 3 int main()
 4 {
 5     freopen("A-large-practice.in", "r", stdin);
 6     freopen("A.out", "w", stdout);
 7     int t, ca=1;
 8     scanf("%d", &t);
 9     while(t--)
10     {
11         printf("Case #%d: ", ca++);
12         int n;
13         scanf("%d", &n);
14         getchar();
15         string ans="";
16         int maxn=0;
17         for(int i=0;i<n;i++)
18         {
19             gets(s);
20             string ss=s;
21             memset(vis, 0, sizeof(vis));
22             int num=0;
23             int len=strlen(s);
24             for(int j=0;j<len;j++)
25                 if(s[j]!=' ' && !vis[s[j]])
26                     num++, vis[s[j]]=1;
27             if(num>maxn)
28                 maxn=num, ans=s;
29             else if(num==maxn)
30                 if(ss<ans)
31                     ans=ss;
32         }
33         cout<<ans<<endl;
34     }
35     return 0;
36 }

A