HDU 4389 X mod f(x)

X mod f(x)

Time Limit: 2000ms

Memory Limit: 32768KB

This problem will be judged on HDU. Original ID: 4389
64-bit integer IO format: %I64d      Java class name: Main

 

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

Source

2012 Multi-University Training Contest 9

 

解题：数位dp,我们可以枚举数字的各位和进行dp,如果数字和是预定的那么多，并且这个数字模其等于0，那么这个数字正是我们要找的

 

#include <bits/stdc++.h>
using namespace std;
int b[20],dp[20][90][90],ans[60];//长度，余数，位和
int dfs(int len,int mod,int left,int sum,bool flag) {
    if(!len) return left == 0 && sum == mod;
    if(flag && dp[len][left][sum] != -1) return dp[len][left][sum];
    int u = flag?9:b[len],ret = 0;
    for(int i = 0; i <= u; ++i)
        ret += dfs(len-1,mod,(left*10 + i)%mod,sum + i,flag||i < u);
    if(flag) dp[len][left][sum] = ret;
    return ret;
}
int solve(int x,int cur) {
    int len = 0;
    while(x) {
        b[++len] = x%10;
        x /= 10;
    }
    return dfs(len,cur,0,0,false);
}
int main() {
    int kase,cs,x[60],y[60];
    scanf("%d",&kase);
    for(cs = 0; cs < kase; ++cs)
        scanf("%d%d",x + cs,y + cs);
    for(int i = 1; i <= 81; ++i) {
        memset(dp,-1,sizeof dp);
        for(int j = 0; j < kase; ++j)
            ans[j] += solve(y[j],i) - solve(x[j]-1,i);
    }
    for(cs = 0; cs < kase; ++cs)
        printf("Case %d: %d
", cs + 1, ans[cs]);
    return 0;
}