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  • 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): 2992    Accepted Submission(s): 1171


    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
     
    Author
    WHU
     
    Source
     
     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstdlib>
     4 #include <cstring>
     5 #include <algorithm>
     6 #include <stack>
     7 #include <queue>
     8 #include <cmath>
     9 #include <map>
    10 #define ll  __int64
    11 #define dazhi 2147483647
    12 #define bug() printf("!!!!!!!")
    13 #define M 100005
    14 using namespace  std;
    15 int bit[10];
    16 int dp[10][82][82][82];
    17 int  n;
    18 int  functi(int pos,int mod,int xx ,int sum,bool flag)
    19 {
    20     if(pos==0) return (xx==sum&&mod%sum==0);
    21     if(flag&&dp[pos][mod][xx][sum]!=-1) return dp[pos][mod][xx][sum];
    22     ll x=flag ? 9 : bit[pos];
    23     ll ans=0;
    24     for(ll i=0;i<=x;i++)
    25     ans+=functi(pos-1,(mod*10+i)%xx,xx,sum+i,flag||i<x);
    26     if(flag)
    27         dp[pos][mod][xx][sum]=ans;
    28     return ans;
    29 }
    30 int  fun(int x)
    31 {
    32     int len=0;
    33     while(x)
    34     {
    35         bit[++len]=x%10;
    36         x/=10;
    37     }
    38     ll re=0;
    39     for(int i=1;i<=81;i++)
    40         re+=functi(len,0,i,0,0);
    41     return re;
    42 }
    43 int main()
    44 {
    45    int t;
    46    int  l,r;
    47    while(scanf("%d",&t)!=EOF)
    48    {
    49        memset(dp,-1,sizeof(dp));
    50        for(int i=1;i<=t;i++)
    51        {
    52            scanf("%d %d",&l,&r);
    53            int exm=fun(r);
    54            printf("Case %d: %d
    ",i,exm-fun(l-1));
    55        }
    56    }
    57     return 0;
    58 }
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  • 原文地址:https://www.cnblogs.com/hsd-/p/6597263.html
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