随笔 - 531 文章 - 0 评论 - 3 阅读 - 10215

Yaroslav and Two Strings CF296B

如果两个只包含数字且长度为

明显的容斥原理

但注意 a[i]==b[i]

TOTAL - ans1- ans2- ans3

// LUOGU_RID: 105563032
#include <iostream>
#include <cmath>
using namespace std;
 const int  N=1e5+4;
 #define int long long
 const int mod= 1e9+7;
 char a[N],b[N];
 int n,f[N],g[N] ;
 int T;
  
 int Cnt=0;
  
 int Pow(int x,int y){
    if(y==0) return 1;
    int t=Pow(x,y/2);
    if(y&1) return ((t*t)%mod*x)%mod;
    return (t*t)%mod;
 }
  
 signed main(){
    int i,j;
    cin>>n;
    for(i=1;i<=n;++i){
        cin>>a[i]; if(a[i]!='?') a[i]-='0';else Cnt++;
    }
    for(i=1;i<=n;++i){
        cin>>b[i]; if(b[i]!='?') b[i]-='0';else Cnt++;
    }
    Cnt=Pow(10,Cnt) ;
     
    f[0]=g[0]=1;
    for(i=0;i<=9;i++)
     for(j=0;j<=i;j++) T++;
     
    for(i=1;i<=n;i++){
        if(a[i]=='?'&&b[i]=='?') f[i]=f[i-1]*T;
        else if(a[i]=='?'&&b[i]!='?') 
            f[i]=f[i-1]*(10-b[i]);
        else if(a[i]!='?'&&b[i]=='?')
            f[i]=f[i-1]*(1+a[i]);
        else
        f[i]=f[i-1]*(a[i]>=b[i]);
        f[i]%=mod;
        if(f[i]==0) break;
    }
    for(i=1;i<=n;i++){
        if(a[i]=='?'&&b[i]=='?') g[i]=g[i-1]*T;
        else if(a[i]=='?'&&b[i]!='?') 
            g[i]=g[i-1]*(1+b[i]);
        else if(a[i]!='?'&&b[i]=='?')
            g[i]=g[i-1]*(10-a[i]);
        else
        g[i]=g[i-1]*(a[i]<=b[i]);
        g[i]%=mod;
        if(g[i]==0) break;
    }
    int tmp =1;
    for(i=1;i<=n&&tmp;i++){
        if(a[i]=='?'&&b[i]=='?') tmp*=10;
        else if(a[i]=='?'&&b[i]!='?') 
            ;
        else if(a[i]!='?'&&b[i]=='?')
            ;
        else tmp*=(a[i]==b[i]);
        tmp%=mod;
        if(tmp==0) break;
    }
  
    cout<<( (Cnt-f[n]-g[n]+tmp)%mod+mod)%mod <<endl;
 }
 
 
 
 
  