HDU 4389(数位dp)

X mod f(x) Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3550    Accepted Submission(s): 1389   Problem Description Here is a function f(x): 　　　int f ( int x ) { 　　　 if ( x == 0 ) return 0; 　　　 return f ( x / 10 ) + x % 10; 　　　} 　　　Now, you want to know, in a given interval [A, B] (1 <= A <= B <= 109), how many integer x that mod f(x) equal to 0.   Input 　　　The first line has an integer T (1 <= T <= 50), indicate the number of test cases. 　　　Each test case has two integers A, B.   Output 　　　For each test case, output only one line containing the case number and an integer indicated the number of x.     Sample Input 2 1 10 11 20   Sample Output Case 1: 10 Case 2: 3

题意: f(x) 即 数x 各个位数之和, 题目要求 求从A到B 满足 x % f(x) == 0 的数的个数

思路: 按数位dp的套路, 需要先 确定一个dp数组,并且这个dp数组的每一个下标只能 确定唯一个数 x

这里dp数组为四维, dp[pos][sum][mod][res], pos位数,此数和为sum,对mod取模,取模得到的就结果为res

AC代码:

#include
#include
using namespace std;

int dp[11][82][82][82];       // 这里的四维占空间较大,改变任意一维的大小都可能引起MLE
int a[15];

int dfs(int pos,int sum,int mod,int res,bool limit){
    if(pos == -1) return res == 0 && sum == mod;
    if(!limit && dp[pos][sum][mod][res] != -1) return dp[pos][sum][mod][res];
    int up = limit ? a[pos] : 9;
    int tmp = 0,sum_x,res_x;
    for(int i = 0;i <= up;i ++){
        sum_x = sum + i;
        res_x = (res * 10 + i) % mod;
        tmp += dfs(pos-1,sum_x,mod,res_x,limit && i == a[pos]);
    }
    if(!limit) dp[pos][sum][mod][res] = tmp;
    return tmp;
}

int slove(int x){
    int pos = 0;
    while(x){
        a[pos ++] = x % 10;
        x /= 10;
    }
    int ans = 0;
    for(int i = 1;i <= 81;i ++)        //枚举 模
        ans += dfs(pos-1,0,i,0,true);
    return ans;
}

int main()
{
    int t; scanf("%d",&t);
    memset(dp,-1,sizeof(dp));
    int f = 1,n,m;
    while(~scanf("%d%d",&n,&m)){
        printf("Case %d: %d\n",f++,slove(m)-slove(n-1));
    }
    return 0;
}