标签：ota 如何 mes form man using space pac div
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1231 Accepted Submission(s): 651
dp[i][j][k][z]：i:处理的数位，j:该数对13取模以后的值，k:是否已经包含13,z结尾的数
#include <iostream> #include <string.h> #include <algorithm> #include <stdio.h> using namespace std; int dp[12][15][2][10]; int bit[12]; int dfs(int pos,int num,bool t,int e,bool flag) { if(pos==-1)return t&&(num==0); if(!flag && dp[pos][num][t][e]!=-1) return dp[pos][num][t][e]; int end=flag?bit[pos]:9; int ans=0; for(int i=0;i<=end;i++) ans+=dfs(pos-1,(num*10+i)%13,t||(e==1&&i==3),i,flag&&(i==end)); if(!flag)dp[pos][num][t][e]=ans; return ans; } int calc(int n) { int pos=0; while(n) { bit[pos++]=n%10; n/=10; } return dfs(pos-1,0,0,0,1); } int main() { //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); int n; memset(dp,-1,sizeof(dp)); while(scanf("%d",&n)==1) { printf("%d\n",calc(n)); } return 0; }
标签：ota 如何 mes form man using space pac div
原文地址：https://www.cnblogs.com/fht-litost/p/8973679.html
