CF-296B-Yaroslav and Two Strings（容斥）
题意：http://codeforces.com/problemset/problem/296/B
有一个位置ai<bi，有一个位置ai>bi的串的数量。
思路：
总数-全小于等于-全大于等于+全相等
1 #define IOS ios_base::sync_with_stdio(0); cin.tie(0); 2 #include <cstdio>//sprintf islower isupper 3 #include <cstdlib>//malloc exit strcat itoa system("cls") 4 #include <iostream>//pair 5 #include <fstream>//freopen("C:\\Users\\13606\\Desktop\\Input.txt","r",stdin); 6 #include <bitset> 7 //#include <map> 8 //#include<unordered_map> 9 #include <vector> 10 #include <stack> 11 #include <set> 12 #include <string.h>//strstr substr strcat 13 #include <string> 14 #include <time.h>// srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9; 15 #include <cmath> 16 #include <deque> 17 #include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less 18 #include <vector>//emplace_back 19 //#include <math.h> 20 #include <cassert> 21 #include <iomanip> 22 //#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor 23 #include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare) 24 using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation 25 //****************** 26 clock_t __START,__END; 27 double __TOTALTIME; 28 void _MS(){__START=clock();} 29 void _ME(){__END=clock();__TOTALTIME=(double)(__END-__START)/CLOCKS_PER_SEC;cout<<"Time: "<<__TOTALTIME<<" s"<<endl;} 30 //*********************** 31 #define rint register int 32 #define fo(a,b,c) for(rint a=b;a<=c;++a) 33 #define fr(a,b,c) for(rint a=b;a>=c;--a) 34 #define mem(a,b) memset(a,b,sizeof(a)) 35 #define pr printf 36 #define sc scanf 37 #define ls rt<<1 38 #define rs rt<<1|1 39 typedef pair<int,int> PII; 40 typedef vector<int> VI; 41 typedef unsigned long long ull; 42 typedef long long ll; 43 typedef double db; 44 const db E=2.718281828; 45 const db PI=acos(-1.0); 46 const ll INF=(1LL<<60); 47 const int inf=(1<<30); 48 const db ESP=1e-9; 49 const int mod=(int)1e9+7; 50 const int N=(int)1e5+10; 51 52 char s[N],ss[N]; 53 ll dp[N][4]; 54 int a[20]={1,2,3,4,5,6,7,8,9,10}; 55 int aa[20]={10,9,8,7,6,5,4,3,2,1}; 56 57 int main() 58 { 59 int n; 60 sc("%d",&n); 61 sc("%s%s",s+1,ss+1); 62 for(int i=1;i<=n;++i) 63 { 64 if(s[i]=='?'&&ss[i]=='?') 65 dp[i][0]=dp[i][1]=55,dp[i][3]=10; 66 else if(s[i]=='?') 67 { 68 dp[i][0]=a[ss[i]-'0']; 69 dp[i][1]=aa[ss[i]-'0']; 70 dp[i][3]=1; 71 } 72 else if(ss[i]=='?') 73 { 74 dp[i][0]=aa[s[i]-'0']; 75 dp[i][1]=a[s[i]-'0']; 76 dp[i][3]=1; 77 } 78 else 79 { 80 if(s[i]<=ss[i]) 81 dp[i][0]=1; 82 if(s[i]>=ss[i]) 83 dp[i][1]=1; 84 if(s[i]==ss[i]) 85 dp[i][3]=1; 86 } 87 } 88 ll ansmin=1,ansmax=1,ansmix=1,anseq=1; 89 for(int i=1;i<=n;++i) 90 { 91 ansmin*=dp[i][0]; 92 ansmax*=dp[i][1]; 93 anseq*=dp[i][3]; 94 if(s[i]=='?'&&ss[i]=='?') 95 ansmix*=100; 96 else if(s[i]=='?'||ss[i]=='?') 97 ansmix*=10; 98 ansmin%=mod; 99 ansmax%=mod; 100 ansmix%=mod; 101 anseq%=mod; 102 } 103 pr("%lld\n",((ansmix-ansmin-ansmax+anseq)%mod+mod)%mod); 104 return 0; 105 } 106 107 /**************************************************************************************/
