Round#179 diy2第二題

   題意：給你兩個數字串，其中可以包含'?'，求將兩個數字串變成不可比較的數字串有多少種方案。（不可比較當且僅當存在i,j,使得s[i]<w[i]&&s[j]>w[j])。

題意理解清楚後，就知道這題其實就是一道排列組合的題目了。

//排列組合
#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;
const int N=100005;
const int M=1000000007;
char s[N],w[N];//輸入字符串
int n;

int main()
{
    int cnt;//'?'個數
    int small,big;
    while(~scanf("%d",&n))
    {
        cnt=0;
        small=big=0;
        scanf("%s%s",s,w);
        for(int i=0;i<n;i++)
        {
            if(s[i]=='?'||w[i]=='?')
            {
                if(s[i]=='?')
                    cnt++;
                if(w[i]=='?')
                    cnt++;
            }
            else
            {
                if(s[i]<w[i])
                    small=1;//s[i]<w[i]
                else if(s[i]>w[i])
                    big=1;//s[i]>w[i]
            }
        }
        if(small==1&&big==1)//除'?'以外的其它字符已經符合題意，則'?'可以隨意替換
        {
            long long ans=1;
            for(int i=1;i<=cnt;i++)
                ans=ans*10%M;
            printf("%d\n",ans);
        }
        else if(cnt==0)//無'?'
            printf("0\n");
        else if(cnt==n*2)//全'?'
        {
            long long sum,tmp1,tmp2;
            sum=tmp1=tmp2=1;
            for(int i=1;i<=cnt;i++)
                sum=sum*10%M;
            for(int i=1;i<=n;i++)
            {
                tmp1=tmp1*55%M;//大於或等於
                tmp2=tmp2*10%M;//等於
            }
            sum=(sum-tmp1*2+tmp2+M*2)%M;
            printf("%d\n",sum);
        }
        else
        {
            if(small)//小於或等於
            {
                long long sum=1,tmp=1;
                for(int i=1;i<=cnt;i++)
                    sum=sum*10%M;
                for(int i=0;i<n;i++)
                {
                    if(s[i]=='?'&&w[i]=='?')
                        tmp=tmp*55%M;
                    else if(s[i]=='?')
                        tmp=tmp*(w[i]-'0'+1)%M;
                    else if(w[i]=='?')
                        tmp=tmp*(10-(s[i]-'0'))%M;
                }
                sum=(sum-tmp+M)%M;
                printf("%d\n",sum);
            }
            else if(big)//大於或等於
            {
                long long sum=1,tmp=1;
                for(int i=1;i<=cnt;i++)
                    sum=sum*10%M;
                for(int i=0;i<n;i++)
                {
                    if(s[i]=='?'&&w[i]=='?')
                        tmp=tmp*55%M;
                    else if(s[i]=='?')
                        tmp=tmp*(10-(w[i]-'0'))%M;
                    else if(w[i]=='?')
                        tmp=tmp*(s[i]-'0'+1)%M;
                }
                sum=(sum-tmp+M)%M;
                printf("%d\n",sum);
            }
            else
            {
                long long sum,tmp1,tmp2,tmp3;
                sum=tmp1=tmp2=tmp3=1;
                for(int i=1;i<=cnt;i++)
                    sum=sum*10%M;
                for(int i=0;i<n;i++)
                {
                    if(s[i]=='?'&&w[i]=='?')
                    {
                        tmp1=tmp1*55%M;
                        tmp2=tmp2*55%M;
                        tmp3=tmp3*10%M;
                    }
                    else if(s[i]=='?')
                    {
                        tmp1=tmp1*(10-(w[i]-'0'))%M;
                        tmp2=tmp2*(w[i]-'0'+1)%M;
                    }
                    else if(w[i]=='?')
                    {
                        tmp1=tmp1*(s[i]-'0'+1)%M;
                        tmp2=tmp2*(10-(s[i]-'0'))%M;
                    }
                }
                sum=(sum-tmp1-tmp2+tmp3+M*2)%M;
                printf("%d\n",sum);
            }
        }
    }
    return 0;
}