Codeforces Round #179 (Div. 2) B.Yaroslav and Two Strings(容斥原理)

题目链接

比赛的时候题给看错了。。当作神题了，看了一下题解，思考了好一会，终于在虎哥的帮助下，明白了。

求出所有的位置上s1[i]>=s2[i]的情况数，s1[i] <= s2[i]的情况数，小小的容斥一下，即可。中间各种错误，各种WA。

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 using namespace std;
 5 #define LL __int64
 6 #define MOD 1000000007
 7 char s1[100001],s2[100001];
 8 int sum[100001],p1[100001],p2[100001],p3[100001];
 9 LL a,b,c,d;
10 int main()
11 {
12     int n,i;
13     scanf("%d%s%s",&n,s1,s2);
14     for(i = 0;i < n;i ++)
15     {
16         if(s1[i] == '?'&&s2[i] == '?')
17         {
18             sum[i] = 100;
19             p1[i] = 55;
20             p2[i] = 55;
21             p3[i] = 10;
22         }
23         else if(s1[i] == '?')
24         {
25             sum[i] = 10;
26             p1[i] = 10 - (s2[i]-'0');
27             p2[i] = s2[i] - '0' + 1;
28             p3[i] = 1;
29         }
30         else if(s2[i] == '?')
31         {
32             sum[i] = 10;
33             p1[i] = s1[i] - '0'+1;
34             p2[i] = 10 - (s1[i]-'0');
35             p3[i] = 1;
36         }
37         else
38         {
39             if(s1[i] >= s2[i])
40             p1[i] = 1;
41             if(s1[i] <= s2[i])
42             p2[i] = 1;
43             if(s1[i] == s2[i])
44             p3[i] = 1;
45             sum[i] = 1;
46         }
47     }
48     a = b = c = d = 1;
49     for(i = 0;i < n;i ++)
50     {
51         a = ((LL)(a*sum[i]))%MOD;
52         b = ((LL)(b*p1[i]))%MOD;
53         c = ((LL)(c*p2[i]))%MOD;
54         d = ((LL)(d*p3[i]))%MOD;
55     }
56     a = (a - b - c + d)%MOD;
57     if(a < 0)
58     a += MOD;
59     printf("%I64d\n",a);
60     return 0;
61 }

 

posted @ 2013-04-16 21:14  Naix_x  阅读(303)  评论(0)  编辑  收藏  举报