hdu4351 Digital root

最新推荐文章于 2019-02-27 15:31:25 发布

magicnumber

最新推荐文章于 2019-02-27 15:31:25 发布

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文章标签： tree integer query each build output

本文链接：https://blog.youkuaiyun.com/magicnumber/article/details/7859973

本文介绍了一种使用线段树和状态压缩技术优化数字根查询的方法。通过预处理和维护序列中数字根的状态，实现了快速查询区间内最大的五个不同的数字根。文章提供了完整的代码实现，并解释了关键步骤。

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Digital root

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 98304/98304 K (Java/Others)
Total Submission(s): 1056 Accepted Submission(s): 328

Problem Description

The digital root (also repeated digital sum) of a number is the (single digit) value obtained by an iterative process of summing digits, on each iteration using the result from the previous iteration to compute a digit sum. The process continues until a single-digit number is reached.
For example, the digital root of 65536 is 7, because 6+5+5+3+6=25 and 2+5=7.
The digital root of an interval is digital root of the sum of all numbers in that interval.

Consider a sequence A ₁,A ₂,A ₃……A _n, your task is to answer some queries[l,r]. For each query, tell the biggest five different digital roots among all continuous subintervals of interval[l,r]

Input

In the first line of the input , we have one integer T indicating the number of test cases. (1 <= T <= 10) Then T cases follows. For each test case, the first line is an integer n indicating the length of sequence(1<=n<=100000); following one line containing n integer A _i(0<=A _i<=1000000000); the third line is an integer Q indicating the number of queries(1<=Q<=100000); following Q lines, each line indicating one query [l,r],(1<=l<=r<=n).

Output

For each test case, first you should print "Case #x:" in a line, where x stands for the case number started with 1. Then for each query output a line contains five integer indicating the answers(if the kind of digital root is smaller than five, output -1 to complete). Leave an empty line between test cases.

Sample Input

  
   1
5
101 240 331 4 52
3
1 3
4 5
1 5

Sample Output

  
   Case #1:
8 7 6 4 2
7 4 2 -1 -1
9 8 7 6 4

   
    
     Hint
    
For the first query, [1,3] has six subintervals: [1,1],[2,2],[3,3],[1,2],[2,3],[1,3]. Interval sums are 101,240,331,341,571,672, correspondence digital roots are 2,6,7
,8,4,6,the most biggest five are 8,7,6,4,2

Author

L7we@UESTC_Alchemist

Source

2012 Multi-University Training Contest 6

Recommend

zhuyuanchen520

好蛋疼的线段树题。

维护前面，中间和后面三个部分。

数字根就是一个数每一位加起来mod9，如果结果是0数字根就是9，不然就是mod9的值，注意判断原来的数是不是就是0.

需要用状态压缩下10个数出现的情况，不然会tle。

代码

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

typedef struct
{
    int l,r,sum;
    int lsum,rsum,midsum;
}Tree;

Tree tree[500005];
int a[100005];
int c[2000][2000];
int ans,rans;

int Count(int t)
{
    if (t==0) return 0;
    int s=0;
    while(t)
    {
        s+=t%10;
        t/=10;
    }
    s%=9;
    if (s==0) return 9;
    return s;
}

int Pre()
{
    int i,j,x,y;
    for (i=0;i<1024;i++)
    {
        for (j=i;j<1024;j++)
        {
            c[i][j]=0;
            for (x=0;x<10;x++)
            {
                if ((i & (1<<x))==0) continue;
                for (y=0;y<10;y++)
                {
                    if ((j & (1<<y))==0) continue;
                    c[i][j]=(c[i][j]|(1<<Count(x+y)));
                }
            }
            c[j][i]=c[i][j];
        }
    }
}

void Build(int t,int l,int r)
{
    tree[t].l=l;
    tree[t].r=r;
    if (l==r)
    {
        tree[t].sum=a[l];
        tree[t].lsum=(1<<a[l]);
        tree[t].rsum=(1<<a[l]);
        tree[t].midsum=(1<<a[l]);
        return;
    }
    int mid=(l+r)/2;
    Build(2*t+1,l,mid);
    Build(2*t+2,mid+1,r);
    tree[t].sum=Count(tree[2*t+1].sum+tree[2*t+2].sum);
    tree[t].lsum=(tree[2*t+1].lsum | c[1<<tree[2*t+1].sum][tree[2*t+2].lsum]);
    tree[t].rsum=(tree[2*t+2].rsum | c[tree[2*t+1].rsum][1<<tree[2*t+2].sum]);
    tree[t].midsum=(tree[2*t+1].midsum | tree[2*t+2].midsum | c[tree[2*t+1].rsum][tree[2*t+2].lsum]);
}

void Query(int t,int x,int y)
{
    int l,r;
    l=tree[t].l;
    r=tree[t].r;
    if (l==x && r==y)
    {
        ans=(ans | tree[t].midsum | c[rans][tree[t].lsum]);
        rans=(c[rans][1<<tree[t].sum] | tree[t].rsum);
        return;
    }
    int mid=(l+r)/2;
    if (mid>=x) Query(2*t+1,x,min(y,mid));
    if (mid<y) Query(2*t+2,max(x,mid+1),y);
}

int main()
{
    int T,i,j,n,x,y,m,s,cnt=1;
    Pre();
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for (i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            a[i]=Count(a[i]);
        }
        Build(0,0,n-1);
        scanf("%d",&m);
        printf("Case #%d:\n",cnt++);
        while(m--)
        {
            scanf("%d%d",&x,&y);
            ans=rans=0;
            Query(0,x-1,y-1);
            s=0;
            for (i=9;i>=0;i--)
            {
                if (ans>=(1<<i))
                {
                    ans-=(1<<i);
                    if (s==0) printf("%d",i);
                    else printf(" %d",i);
                    s++;
                    if (s==5) break;
                }
            }
            for (i=s;i<5;i++)
            {
                printf(" -1");
            }
            printf("\n");
        }
        if (T>0) printf("\n");
    }
    return 0;
}