str2int
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1568 Accepted Submission(s): 540
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.
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.
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
101
123
09
000
1234567890
Sample Output
202
Source
题意: n个字符串，对于每一个子串可以表示为一个数字， 求所有子串的数字和相加对2012取模，， 相同数字只算一次。
这题可以先把n个字符串用一个没有出现过的字符隔开连起来。然后求sa， lcp。
我们可以先看一个简单的例子。
s = 12345
num[1] = 1 sum[1] = 1
num[2] = 12 sum[2] = 1 + 12
num[3] = 123 sum[3] = 1 + 12 + 123
num[4] = 1234 sum[4] = 1 + 12 ＋ 123 + 1234
num[5] = 12345 sum[5] = 1 + 12 + 123 + 1234 + 12345
如果求[3, 4] ， 只需要 sum[5] - sum[2] - num[2] * (10 + 100 + 1000)；
判重时 只要从 i+ lcp[rank[i]] 开始算就可以了，，因为公共前缀那一部分 在前面已经算了。
上代码。。
#include <set>
#include <map>
#include <cmath>
#include <ctime>
#include <queue>
#include <stack>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const double eps = 1e-;
const int mod = ;
const int maxn = 2e5+;
int sum [maxn], num[maxn];
string s;
int sa[maxn], Rank[maxn], tmp[maxn], lcp[maxn];
int k, len;
bool cmp(int i, int j)
{
if (Rank[i] != Rank[j])
return Rank[i] < Rank[j];
else
{
int x = (i+k <= len ? Rank[i+k] : -);
int y = (j+k <= len ? Rank[j+k] : -);
return x < y;
}
}
void build_sa()
{
for (int i = ; i <= len; i++)
{
sa[i] = i;
Rank[i] = (i < len ? s[i] : -);
}
for (k = ; k <= len; k *= )
{
sort (sa,sa+len+,cmp);
tmp[sa[]] = ;
for (int i = ; i <= len; i++)
{
tmp[sa[i]] = tmp[sa[i-]] + (cmp(sa[i-],sa[i])? : );
}
for (int i = ; i <= len; i++)
Rank[i] = tmp[i];
}
} void Get_lcp()
{
for (int i = ; i < len; i++)
Rank[sa[i]] = i;
int h = ;
lcp[] = ;
for (int i = ; i < len; i++)
{
int j = sa[Rank[i]-];
if (h > )
h--;
for (; h+i < len && h+j < len; h++)
if (s[i+h] != s[j+h])
break;
lcp[Rank[i]] = h;
}
}
bool isdigit(char &ch)
{
return ch >= '' && ch <= '';
}
int vec[maxn], board[maxn], tot;
int SUM[maxn];
int solve (int l, int r)
{
if (l > r)
return ;
int res ;
res = sum[r] - sum[l-];
res = ((res%mod)+mod)%mod;
res -= num[l-] * SUM[r-l+];
res = ((res%mod)+mod)%mod;
return ((res%mod)+mod)%mod;
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
#endif
int n;
SUM[] = ;
for (int i = ; i < maxn; i++)
{
SUM[i] = (SUM[i-] + ) * % mod;
}
while (~scanf ("%d", &n))
{
s = "\0";
tot = ;
memset(sum, , sizeof(sum));
memset(num, , sizeof(num));
for (int i = ; i < n; i++)
{
string tmp;
cin >> tmp;
s += tmp + "#";
}
len = s.size();
int val = ;
for (int i = ; i < len; i++)
{
if (s[i] != '#')
{
val = (val * + s[i] - '') % mod;
num[i] = val;
sum[i] = (sum[i-] + num[i]) % mod;
board[i] = tot;
}
if (s[i] == '#')
{
vec[tot++] = i;
num[i] = val;
sum[i] = sum[i-] + val;
}
}
build_sa();
Get_lcp();
int ans = ;
for (int i = ; i < len; i++)
{
int t1 = i + lcp[Rank[i]];
if (s[i] == '')
continue;
if (isdigit(s[i]) && i+lcp[Rank[i]] < vec[board[i]])
{
int t2 = vec[board[i]] -;
int ans1 = solve(i, t2);
int ans2 = solve(i , t1-);
ans = (ans + solve(i, t2) - solve(i, t1-)) % mod;
if (ans < )
ans += mod;
}
}
printf("%d\n", ans%mod);
}
return ;
}