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CF-296B-Yaroslav and Two Strings(容斥)

时间:2020-03-29 20:33:19      阅读:64      评论:0      收藏:0      [点我收藏+]

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 /**************************************************************************************/

 

CF-296B-Yaroslav and Two Strings(容斥)

原文:https://www.cnblogs.com/--HPY-7m/p/12594461.html

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