-
数位dp模板
- typedef long long ll;
- int a[20];
- ll dp[20][state];
- ll dfs(int pos,
- {
-
- if(pos==-1) return 1;
-
- if(!limit && !lead && dp[pos][state]!=-1) return dp[pos][state];
-
- int up=limit?a[pos]:9;
- ll ans=0;
-
- for(int i=0;i<=up;i++)
- {
- if() ...
- else if()...
- ans+=dfs(pos-1,
-
- }
-
- if(!limit && !lead) dp[pos][state]=ans;
-
- return ans;
- }
- ll solve(ll x)
- {
- int pos=0;
- while(x)
- {
- a[pos++]=x%10;
- x/=10;
- }
- return dfs(pos-1
- }
- int main()
- {
- ll le,ri;
- while(~scanf("%lld%lld",&le,&ri))
- {
-
- printf("%lld
",solve(ri)-solve(le-1));
- }
- }
-
相关阅读:
序列化与反序列化
进程与线程
winform基础
MD5加密
Docker安装Nextcloud+collabora office+ocdownloader
Docker安装MariaDB
Docker 安装 Nginx
Docker命令大全
Docker之镜像操作
Linux入门-Docker安装
-
原文地址:https://www.cnblogs.com/renwjing/p/7325283.html
Copyright © 2020-2023
润新知