zoukankan      html  css  js  c++  java
  • hdu4389 X mod f(x)[数位dp]

    X mod f(x)

    Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 3020    Accepted Submission(s): 1182


    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
     
    Recommend
    zhuyuanchen520   |   We have carefully selected several similar problems for you:  4385 4383 4388 4387 4386 
     

    由于f(x)最大就是81,所以可以算对于1-81每一个数都求一下就可以

    #include<cstdio>
    #include<cstring>
    using namespace std;
    int cas,T,bits[20];
    int f[11][82][82][82];
    int dfs(int pos,int mod,int x,int sum,bool lim){
        if(!pos){return x==sum&&!mod;}
        int &res=f[pos][mod][x][sum],ans=0;
        if(!lim&&(~res)) return res;
        int up=!lim?9:bits[pos];
        for(int i=0;i<=up;i++){
            ans+=dfs(pos-1,(mod*10+i)%x,x,sum+i,lim&&i==bits[pos]);
        }
        if(!lim) res=ans;
        return ans;
    }
    int solve(int x){
        int len=0;int ans=0;
        for(;x;x/=10) bits[++len]=x%10;
        for(int fx=1;fx<=81;fx++) ans+=dfs(len,0,fx,0,1);
        return ans;
    }
    int main(){
        int l,r;
        memset(f,-1,sizeof f);
        for(scanf("%d",&T);T--;){
            scanf("%d%d",&l,&r);
            printf("Case %d: %d
    ",++cas,solve(r)-solve(l-1));
        }
        return 0;
    }
  • 相关阅读:
    052-141
    052-140
    052-139
    052-138
    需要做笔记的页面
    日期总是显示1900/01/01 0:00:00
    延迟加载的树控件
    (简单)关于summary的注释
    江南检测
    fineui动态添加用户控件
  • 原文地址:https://www.cnblogs.com/shenben/p/6708043.html
Copyright © 2011-2022 走看看