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  • 数位DP

    #include<bits/stdc++.h>
    #define N 22
    #define M 11
    #define db double
    #define ll long long
    #define ldb long double
    #define ull unsigned long long
    using namespace std;
    const int h=3,ki=149,mo=998244353;
    int mod(int x){return (x%mo+mo)%mo;}
    int inc(int x,int k){x+=k;return x<mo?x:x-mo;}
    int dec(int x,int k){x-=k;return x>=0?x:x+mo;}
    int ksm(int x,int k)
    {
    	int ans=1;
    	while(k){if(k&1)ans=1ll*ans*x%mo;k>>=1;x=1ll*x*x%mo;}
    	return mod(ans);
    }
    int inv(int x){return ksm(x,mo-2);}
    int read()
    {
    	char ch=0;int x=0,flag=1;
    	while(!isdigit(ch)){ch=getchar();if(ch=='-')flag=-1;}
    	while(isdigit(ch)){x=(x<<3)+(x<<1)+ch-'0',ch=getchar();}
    	return x*flag;
    }
    void write(int x)
    {
    	if(!x)return (void)putchar(48);
    	if(x<0)putchar(45),x=-x;
    	int len=0,p[20];
    	while(x)p[++len]=x%10,x/=10;
    	for(int i=len;i>=1;i--)putchar(p[i]+48);
    }
    int a[N],dp[N][M][2];
    int solve(int s)
    {
    	if(s==0)return 0;
    	int n=0;while(s)a[++n]=s%10,s/=10;reverse(a+1,a+n+1);
    	
    	memset(dp,0,sizeof(dp));
    	for(int i=1;i<=a[1];i++)dp[1][i][i==a[1]]=1;
    	for(int x=2;x<=n;x++)for(int i=1;i<=9;i++)dp[x][i][0]=1;
    	
    	for(int x=1;x<n;x++)
    	for(int lst=0;lst<=9;lst++)
    	if(dp[x][lst][0]||dp[x][lst][1])
    	for(int i=0;i<=9;i++)if(abs(i-lst)>=2)
    	{
    		int f0=dp[x][lst][0];
    		int f1=dp[x][lst][1];
    		int &g0=dp[x+1][i][0];
    		int &g1=dp[x+1][i][1];
    		
    		if(i<a[x+1])g0+=f0+f1;
    		if(i>=a[x+1])g0+=f0;
    		if(i==a[x+1])g1+=f1;
    	}
    	
    	int ans=0;
    	for(int lst=0;lst<=9;lst++)ans+=dp[n][lst][0]+dp[n][lst][1];
    	return ans;
    }
    int main()
    {
    	int l=read(),r=read();
    	write(solve(r)-solve(l-1));
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/Creed-qwq/p/15033049.html
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