codeforces 296B - Yaroslav and Two Strings (DP+容斥)

最新推荐文章于 2021-06-02 15:12:03 发布

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最新推荐文章于 2021-06-02 15:12:03 发布

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分类专栏：其他DP 容斥君

本文链接：https://blog.youkuaiyun.com/baidu_35522092/article/details/74357202

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本文介绍了一种使用动态规划算法解决字符串比较问题的方法，旨在计算两个由数字和问号组成的字符串模板，在替换问号后的非可比性组合数。通过详细解析算法流程，帮助读者理解如何高效地解决此类问题。

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Yaroslav thinks that two strings s and w, consisting of digits and having length n are non-comparable if there are two numbers, i and j (1 ≤ i, j ≤ n), such that s_i > w_i and s_j < w_j. Here sign s_i represents the i-th digit of string s, similarly, w_j represents the j-th digit of string w.

A string's template is a string that consists of digits and question marks ("?").

Yaroslav has two string templates, each of them has length n. Yaroslav wants to count the number of ways to replace all question marks by some integers in both templates, so as to make the resulting strings incomparable. Note that the obtained strings can contain leading zeroes and that distinct question marks can be replaced by distinct or the same integers.

Help Yaroslav, calculate the remainder after dividing the described number of ways by 1000000007 (10⁹ + 7).

Input

The first line contains integer n (1 ≤ n ≤ 10⁵) — the length of both templates. The second line contains the first template — a string that consists of digits and characters "?". The string's length equals n. The third line contains the second template in the same format.

Output

In a single line print the remainder after dividing the answer to the problem by number 1000000007 (10⁹ + 7).

Examples

Input

2
90
09

Output

Input

2
11
55

Output

Input

5
?????
?????

Output

993531194

Note

The first test contains no question marks and both strings are incomparable, so the answer is 1.

The second test has no question marks, but the given strings are comparable, so the answer is 0.

dp[3][MAXN]:dp[0][i]表示第i位为止所有的A[j]>=B[j]的数目

dp[1][i]表示第i位为止所有的A[j]==B[j]的数目

dp[2][i]表示第i位为止所有的A[j]<=B[j]的数目

ans：A，B中共有多少n个’？’，则ans=10^n;表示A，B共有ans种不同的组合方式（不管满不满足题目要求）

然后ans=ans-dp[0][n]-dp[2][n]+dp[1][n];（先减dp[0][n]保证不会全部大于等于，一定存在小于，再减dp[2][n]保证不会全部小于等于，一定存在大于，此时全部等于的情况减了两次，所以再加上dp[1][n]）

#include <iostream>
#include <algorithm>
#include <vector>
#include <stack>
#include <string.h>
#include <cstdio>
#include <map>
#include <queue>
#include <math.h>
#include <cstring>
#include <set>
#include <hash_map>
#define pb push_back
#define fi first
#define se second
#define INF 0x3f3f3f3f
using namespace std;
typedef long long LL;
const LL MOD=1e9+7;
const
LL MAXN=1e5+23;
LL dp[3][MAXN];//> = <
char a[MAXN],b[MAXN];
int n;

int main()
{
    while(scanf("%d",&n)!=-1)
    {
        scanf("%s%s",a,b);
        if(n==1){puts("0");continue;}
        memset(dp,0,sizeof(dp));
        dp[0][0]=dp[1][0]=dp[2][0]=1;
        LL sum=0,ans=1;
        for(int i=0;i<n;i++)
        {
            if(a[i]=='?')sum++;if(b[i]=='?')sum++;
            if(a[i]=='?')
            {
                if(b[i]=='?')
                {
                    dp[0][i+1]=dp[0][i]*55%MOD;
                    dp[1][i+1]=dp[1][i]*10%MOD;
                    dp[2][i+1]=dp[2][i]*55%MOD;
                }
                else
                {
                    int B=b[i]-'0';
                    dp[0][i+1]=dp[0][i]*(10-B)%MOD;
                    dp[1][i+1]=dp[1][i]*1%MOD;
                    dp[2][i+1]=dp[2][i]*(B+1)%MOD;
                }
            }
            else
            {
                if(b[i]=='?')
                {
                    int A=a[i]-'0';
                    dp[0][i+1]=dp[0][i]*(A+1)%MOD;
                    dp[1][i+1]=dp[1][i]*1%MOD;
                    dp[2][i+1]=dp[2][i]*(10-A)%MOD;
                }
                else
                {
                    int A=a[i]-'0';
                    int B=b[i]-'0';
                    //printf("A=%d,B=%d\n",A,B);
                    if(A==B)
                    {
                        dp[0][i+1]=dp[0][i]*1;
                        dp[1][i+1]=dp[1][i]*1%MOD;
                        dp[2][i+1]=dp[2][i]*1;
                    }
                    else if(A<B)
                    {
                        dp[0][i+1]=0;
                        dp[1][i+1]=0;
                        dp[2][i+1]=dp[2][i]*1%MOD;
                    }
                    else if(A>B)
                    {
                        dp[0][i+1]=dp[0][i]*1%MOD;
                        dp[1][i+1]=0;
                        dp[2][i+1]=0;
                    }
                }
            }
        }
        for(int i=0;i<sum;i++)ans=ans*10%MOD;
        //printf("%lld %lld %lld ans=%lld\n",dp[0][n],dp[1][n],dp[2][n],ans);
        printf("%lld\n",(ans-dp[0][n]-dp[2][n]+dp[1][n]+4*MOD)%MOD);
    }
    return 0;
}

本人蒟蒻，如有错误，还望指正