hdu 4436 str2int（后缀数组）

str2int

Problem Description

In this problem, you are given several strings that contain only digits from '0' to '9', inclusive.
An example is shown below.
101
123
The set S of strings is consists of the N strings given in the input file, and all the possible substrings of each one of them.
It's boring to manipulate strings, so you decide to convert strings in S into integers.
You can convert a string that contains only digits into a decimal integer, for example, you can convert "101" into 101, "01" into 1, et al.
If an integer occurs multiple times, you only keep one of them. 
For example, in the example shown above, all the integers are 1, 10, 101, 2, 3, 12, 23, 123.
Your task is to calculate the remainder of the sum of all the integers you get divided by 2012.

 

Input

There are no more than 20 test cases.
The test case starts by a line contains an positive integer N.
Next N lines each contains a string consists of one or more digits.
It's guaranteed that 1≤N≤10000 and the sum of the length of all the strings ≤100000.
The input is terminated by EOF.

 

Output

An integer between 0 and 2011, inclusive, for each test case.

 

Sample Input

   
   
   
   

    
    
    
    5
101
123
09
000 
1234567890
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    202
   
   
   
   

 

Source

2012 Asia Tianjin Regional Contest

 

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zhoujiaqi2010   |   We have carefully selected several similar problems for you:   5634  5633  5632  5629  5628 

如1234   T[4]=1+12+123+1234  V[2]=12   计算3+34 = T[4]-T[2]-V[2]*110 即可。

#include<cstdio>
#include<cstdlib>
#include<cstring>
#define N 300000
#define MOD 2012
using namespace std;
int wa[N],wb[N],wv[N];
int rank[N],height[N],ws[N],sa[N],P[N],pow[N],V[N],
T[N],Len[N],S[N],arrive[N],n,len,value;
char s[N];
int cmp(int *r,int a,int b,int l)
{
    return r[a]==r[b]&&r[a+l]==r[b+l];
}
void da(int *r,int *sa,int n,int m)
{
    int i,j,p,*x=wa,*y=wb,*t;
    for(i=0;i<m;i++) ws[i]=0;
    for(i=0;i<n;i++) ws[x[i]=r[i]]++;
    for(i=1;i<m;i++) ws[i]+=ws[i-1];
    for(i=n-1;i>=0;i--) sa[--ws[x[i]]]=i;
    for(p=1,j=1;p<n;j*=2,m=p)
    {
        for(p=0,i=n-j;i<n;i++) y[p++]=i;
        for(i=0;i<n;i++) if(sa[i]>=j) y[p++]=sa[i]-j;
        for(i=0;i<n;i++) wv[i]=x[y[i]];
        for(i=0;i<m;i++) ws[i]=0;
        for(i=0;i<n;i++) ws[wv[i]]++;
        for(i=1;i<m;i++) ws[i]+=ws[i-1];
        for(i=n-1;i>=0;i--) sa[--ws[wv[i]]]=y[i];
        for(t=x,x=y,y=t,p=1,x[sa[0]]=0,i=1;i<n;i++)
            x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++;
    }
}
void calheight(int *r,int *sa,int n)
{
    int i,j,k=0;
    for(i=0;i<n;i++) rank[sa[i]]=i;
    for(i=0;i<n;height[rank[i++]]=k)
    if(rank[i])for(k?k--:0,j=sa[rank[i]-1];r[i+k]==r[j+k];k++);
    return;
}
int get(int l,int r)
{
    if(l>r) return 0;
    int ans=(T[r+1]-T[l])%MOD;
    ans-=V[l]*(pow[r-l+1]);
    ans%=MOD;
    if(ans<0) ans+=MOD;
    return ans;
}
int main()
{
    freopen("C:\\Users\\Administrator\\Desktop\\input.txt","r",stdin);
    pow[0]=1;pow[1]=10;
    for(int i=2;i<200000;i++)
        pow[i]=(pow[i-1]+1)*10%MOD;
    while(scanf("%d",&n)!=EOF)
    {
        len=0;value=0;
        for(int i=0;i<n;i++)
        {
            scanf("%s",s);
            Len[i]=strlen(s);
            for(int j=0;j<Len[i];j++)
            {
                S[len]=s[j]-'0'+1;
                value=(value*10+S[len]-1)%MOD;
                V[len+1]=value;
                T[len+1]=(T[len]+value)%MOD;
                P[len]=i;
                len++;
            }
            S[len]=11;//设置分隔符
            value=(value*10+10)%MOD;
            V[len+1]=value;
            T[len+1]=(T[len]+value)%MOD;
            P[len]=i;
            len++;
            arrive[i]=len-2;
        }
        S[len-1]=0;//增加末尾的0
        da(S,sa,len,12);
        calheight(S,sa,len);
        int ans=0;
        for(int i=0;i<len;i++)
        {
            if(S[i]%10!=1)
            {
                if(i+height[rank[i]]<=arrive[P[i]])
                {
                   ans+=get(i,arrive[P[i]])-get(i,i+height[rank[i]]-1);//字符串的值减去下面重复的值
                   ans%=MOD;
                   if(ans<0)  ans+=MOD;
               }
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}