Codecraft-17 and Codeforces Round #391 (Div. 1 + Div. 2, combined)D. Felicity's Big Secret Revealed

题目连接：http://codeforces.com/contest/757/problem/D

D. Felicity's Big Secret Revealed

time limit per test

4 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

The gym leaders were fascinated by the evolutions which took place at Felicity camp. So, they were curious to know about the secret behind evolving Pokemon.

The organizers of the camp gave the gym leaders a PokeBlock, a sequence of n ingredients. Each ingredient can be of type 0 or 1. Now the organizers told the gym leaders that to evolve a Pokemon of type k (k ≥ 2), they need to make a valid set of k cuts on the PokeBlock to get smaller blocks.

Suppose the given PokeBlock sequence is b0b1b2... bn - 1. You have a choice of making cuts at n + 1 places, i.e., Before b0, between b0and b1, between b1 and b2, ..., between bn - 2 and bn - 1, and after bn - 1.

The n + 1 choices of making cuts are as follows (where a | denotes a possible cut):

| b0 | b1 | b2 | ... | bn - 2 | bn - 1 |

Consider a sequence of k cuts. Now each pair of consecutive cuts will contain a binary string between them, formed from the ingredient types. The ingredients before the first cut and after the last cut are wasted, which is to say they are not considered. So there will be exactly k - 1 such binary substrings. Every substring can be read as a binary number. Let m be the maximum number out of the obtained numbers. If all the obtained numbers are positive and the set of the obtained numbers contains all integers from 1 to m, then this set of cuts is said to be a valid set of cuts.

For example, suppose the given PokeBlock sequence is 101101001110 and we made 5 cuts in the following way:

10 | 11 | 010 | 01 | 1 | 10

So the 4 binary substrings obtained are: 11, 010, 01 and 1, which correspond to the numbers 3, 2, 1 and 1 respectively. Here m = 3, as it is the maximum value among the obtained numbers. And all the obtained numbers are positive and we have obtained all integers from 1 to m. Hence this set of cuts is a valid set of 5 cuts.

A Pokemon of type k will evolve only if the PokeBlock is cut using a valid set of k cuts. There can be many valid sets of the same size. Two valid sets of k cuts are considered different if there is a cut in one set which is not there in the other set.

Let f(k) denote the number of valid sets of k cuts. Find the value of . Since the value of s can be very large, output smodulo 109 + 7.

Input

The input consists of two lines. The first line consists an integer n (1 ≤ n ≤ 75) — the length of the PokeBlock. The next line contains the PokeBlock, a binary string of length n.

Output

Output a single integer, containing the answer to the problem, i.e., the value of s modulo 109 + 7.

Examples

input

4
1011

output

10

input

2
10

output

1

Note

In the first sample, the sets of valid cuts are:

Size 2: |1|011, 1|01|1, 10|1|1, 101|1|.

Size 3: |1|01|1, |10|1|1, 10|1|1|, 1|01|1|.

Size 4: |10|1|1|, |1|01|1|.

Hence, f(2) = 4, f(3) = 4 and f(4) = 2. So, the value of s = 10.

In the second sample, the set of valid cuts is:

Size 2: |1|0.

Hence, f(2) = 1 and f(3) = 0. So, the value of s = 1.

题意：给你一个长度为N的01字符串（N<=75），对字符串进行划分，要使得划分的每一部分转换为十进制数出现了一到m(m为转换的最大值)、

题解：dp[i][j]表示在第i个字符结尾j状态的方案数（j表示的状态是j转换成二进制第k位为1的话表示前面的i划分出现过k这个值）

转移方程为dp[k][j|1<<(x-1)]=∑dp[i][j](x为i到k的字符串转换为十进制的那个数)由于字符串最大长度为75则x的最大值为20；然后对答案就是

dp[i][j](0《i《n,j=((1<<k)-1)(1<=k<=20))的和

#include<cstdio> 
#include<algorithm>
#define ll long long
using namespace std;
const int N=80; 
 
const int mod=1e9+7; 
int n,a[N];
char b[N]; 
int dp[N][(1<<20)+10]; 
int main()
{
    scanf("%d",&n);
    scanf("%s",b+1);
    for(int i=1;i<=n;i++)
    {
        a[i]=b[i]-'0'; 
    }
    for(int i=0;i<=n;i++)
    {
        dp[i][0]=1;
        for(int j=0;j<(1<<20);j++)
        {
            if(dp[i][j])
            {
                ll x=0;
                for(int k=i+1;k<=n;k++)
                {
                    x+=a[k];
                    if(x>20)break;
                    if(!x)continue;
                    dp[k][j|1<<(x-1)]=(dp[k][j|1<<(x-1)]+dp[i][j])%mod;    
                    x*=2; 
                } 
            } 
        } 
    } 
    int ans=0; 
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=20;j++)
        {
            ans=(ans+dp[i][(1<<j)-1])%mod; 
        } 
    }
    printf("%d
",ans); 
}