hdu3652

Posted 2020-11-10 lcezfzy

http://acm.hdu.edu.cn/showproblem.php?pid=3652

就是个朴素的数位dp啊，但是我一直tle，毫无办法，来日再战吧。

#include<cstdio>
#include<cstring>
int dp[11][2][2][14];
int num[11];
int remain[11];
int n;
int dfs(int i,bool state,bool have,int k,bool limit) {
    if(i==0){
        if(k==0&&have)
            return 1;
        else return 0;    
    }
    if(~dp[i][state][have][k]&&!limit) return dp[i][state][have][k];
    int up=limit?num[i]:9;
    int res=0;
    for(int j=0;j<=up;j++) {
        int h=(k+j*remain[i])%13;
        res+=dfs(i-1,j==1,have||(state&&j==3),h,limit&&j==up);
    }
    if(!limit) dp[i][state][have][k]=res;
    return res;
}
int solve(int x) {
    memset(dp,-1,sizeof(dp));    
    int len=0;
    while(x) {
        len++;
        num[len]=x%10;
        x/=10;
    }
    remain[1]=1;
    for(int i=2;i<=len;i++) {
        remain[i]=remain[i-1]*10;
    }
    return dfs(len,0,0,0,1);
}
int main() {
    while(scanf("%d",&n)) {
        printf("%d\n",solve(n));
    }
    return 0;
}