zoukankan      html  css  js  c++  java
  • Hdu 4389 X mod f(x) (数位dp)

    X mod f(x)



    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
     
    f(x)是x各位上的数的和,求一个范围内,有多少x使得x mod f(x) == 0。
     
    f(x)显然最多只有81,遍历81种值就可以来数位dp了。
     
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <iostream>
    using namespace std;
    #define ll long long
    int n;
    int dp[10][81][81][81];
    int dig[10];
    
    int dfs(int pos, int num, int mod, int sum, int lim) {
        if(pos == -1) return sum == mod && num == 0;
        if(!lim && dp[pos][sum][mod][num] != -1) return dp[pos][sum][mod][num];
        int End = lim ? dig[pos] : 9;
        int ret = 0;
        for(int i = 0; i <= End; i++) {
            ret += dfs(pos - 1, (num * 10 + i) % mod, mod, sum + i, lim && (i == End));
        }
        if(!lim) dp[pos][sum][mod][num] = ret;
        return ret;
    }
    
    int func(int num) {
        int n = 0;
        while(num) {
            dig[n++] = num % 10;
            num /= 10;
        }
        int ret = 0;
        for(int i = 1; i <= 81; i++)
            ret += dfs(n - 1, 0, i, 0, 1);
        return ret;
    }
    
    int main() {
        int t;
        int val = 0;
        scanf("%d", &t);
        memset(dp, -1, sizeof(dp)); // sb le
        while(t--) {
            int a, b;
            scanf("%d %d", &a, &b);
            printf("Case %d: %d
    ", ++val, func(b) - func(a - 1));
        }
    }
  • 相关阅读:
    .NET平台下,初步认识AutoMapper
    python 二分查找算法
    01背包问题(动态规划)python实现
    NSSM安装服务
    iis .apk .ipa下载设置
    动态规划 转载
    leetcode 5 查找最长的回文子串
    [DEncrypt] MySecurity--安全加密/Base64/文件加密 (转载)
    [DEncrypt] HashEncode--哈希加密帮助类 (转载)
    [DEncrypt] Encrypt--加密/解密/MD5加密 (转载)
  • 原文地址:https://www.cnblogs.com/lonewanderer/p/5661645.html
Copyright © 2011-2022 走看看