Codeforces Round #179 (Div. 2) B. Yaroslav and Two Strings(组合数学)

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).

Sample test(s)

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.

排除不符合条件的替换方法。

AC Code：

  1 #include <iostream>
  2 #include <cstdio>
  3 
  4 using namespace std;
  5 
  6 const int maxn = 100005;
  7 const long long r = 1000000007;
  8 char s[maxn], w[maxn];
  9 int n;
 10 
 11 long long power(long long a, int b)  //计算a^b
 12 {
 13     long long res;
 14     if(b == 0) res = 1;
 15     else if(b == 1) res = a;
 16     else
 17     {
 18         res = power(a, b >> 1);
 19         res = (res * res) % r;
 20         if(b & 1) res = (res * a) % r;
 21         //cout << res << endl;
 22     }
 23     return res;
 24 }
 25 
 26 int main()
 27 {
 28     while(scanf("%d", &n) != EOF)
 29     {
 30         int a = 0, b = 0, c = 0;  //a指示是否存在s[i]>w[i],b指示是否存在s[i]<w[i]
 31         long long ans;
 32         scanf("%s %s", s, w);
 33         if(n == 1)
 34         {
 35             puts("0");
 36             continue;
 37         }
 38         for(int i = 0; i < n; i++)
 39         {
 40             if(s[i] == '?' || w[i] == '?')
 41             {
 42                 if(s[i] == '?') c++;
 43                 if(w[i] == '?') c++;
 44             }
 45             else if(s[i] > w[i]) a = 1;
 46             else if(s[i] < w[i]) b = 1;
 47         }
 48         if(!c)  //全为数字
 49         {
 50             if(a && b) ans = 1;
 51             else ans = 0;
 52         }
 53         else  //有问号
 54         {
 55             ans = power(10, c);
 56             if(a && !b)  //s[i]>w[i]
 57             {
 58                 long long dec = 1;
 59                 for(int i = 0; i < n; i++)
 60                 {
 61                     if(s[i] == '?')
 62                     {
 63                         if(w[i] == '?') dec = (dec * 55) % r;
 64                         else dec = (dec * ('9' - w[i] + 1)) % r;
 65                     }
 66                     else if(w[i] == '?') dec = (dec * (s[i] - '0' + 1)) % r;
 67                 }
 68                 ans -= dec;
 69                 if(ans < 0) ans += r;
 70             }
 71             else if(!a && b)  //s[i]<w[i]
 72             {
 73                 long long dec = 1;
 74                 for(int i = 0; i < n; i++)
 75                 {
 76                     if(w[i] == '?')
 77                     {
 78                         if(s[i] == '?') dec = (dec * 55) % r;
 79                         else dec = (dec * ('9' - s[i] + 1)) % r;
 80                     }
 81                     else if(s[i] == '?') dec = (dec * (w[i] - '0' + 1)) % r;
 82                 }
 83                 ans -= dec;
 84                 if(ans < 0) ans += r;
 85             }
 86             else if(c == 2 * n) //只有?
 87             {
 88                 ans = ans - (2 * power(55, n)) % r + power(10, n);
 89                 if(ans < 0) ans += r;
 90             }
 91             else if(!a && !b )
 92             {
 93                 long long dec1 = 1, dec2 = 1;
 94                 int d = 0;
 95                 for(int i = 0; i < n; i++)
 96                 {
 97                     if(s[i] == '?')
 98                     {
 99                         if(w[i] == '?')
100                         {
101                             dec1 = (dec1 * 55) % r;
102                             d++;
103                         }
104                         else dec1 = (dec1 * ('9' - w[i] + 1)) % r;
105                     }
106                     else if(w[i] == '?') dec1 = (dec1 * (s[i] - '0' + 1)) % r;
107                 }
108                 for(int i = 0; i < n; i++)
109                 {
110                     if(w[i] == '?')
111                     {
112                         if(s[i] == '?') dec2 = (dec2 * 55) % r;
113                         else dec2 = (dec2 * ('9' - s[i] + 1)) % r;
114                     }
115                     else if(s[i] == '?') dec2 = (dec2 * (w[i] - '0' + 1)) % r;
116                 }
117                 ans = ans - dec1 - dec2 + power(10, d);
118                 if(ans < 0) ans += r;
119             }
120         }
121         printf("%I64d\n", ans);
122     }
123     return 0;
124 }

相关阅读:
如何查看MySQL执行的每条SQL
最简单的方式在linux上升级node.js版本
快速理解字符串和编码
macaca常见错误排查
macaca自动化初体验
F2eTest程序快捷方式安装到桌面
centos下mysqlreport安装和使用
前端纯css 图片的模糊处理
gulp入门学习教程（入门学习记录）
关于nodejs中npm命令没有反应的解决方法

原文地址：https://www.cnblogs.com/cszlg/p/3050961.html