标签:author sample time while set scanf mis class key
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 39220 Accepted Submission(s): 14260
/* 数位dp入门题目 一个很好的blog:http://blog.csdn.net/wust_zzwh/article/details/52100392 */ #include<cstdio> #include<cstring> using namespace std; const int N=20; int le,ri,a[N],dp[N][2]; int dfs(int pos,int pre,bool sta,bool limit){ if(!pos) return 1; if(!limit && dp[pos][sta]!=-1) return dp[pos][sta]; int up=limit?a[pos]:9; int ans=0; for(int i=0;i<=up;i++){ if(pre==6 && i==2) continue; if(i==4) continue; ans+=dfs(pos-1,i,i==6,limit && i==a[pos]); } if(!limit) dp[pos][sta]=ans; return ans; } int solve(int x){ int pos=0; for(;x;x/=10) a[++pos]=x%10; return dfs(pos,0,0,1); } int main(){ while(~scanf("%d%d",&le,&ri)&&le&&ri){ memset(dp,-1,sizeof dp); printf("%d\n",solve(ri)-solve(le-1)); } return 0; }
标签:author sample time while set scanf mis class key
原文地址:http://www.cnblogs.com/shenben/p/6411396.html