uva 11361 - Investigating Div-Sum Property(数位dp)

题意:题目大意:给出a,b,k,问说在[a,b]这个区间有多少n,满足n整除k,以及n的各个为上的数字之和也整除k。

分析:dp[i][nmod][smod]长度为i,该数对k的余数,各位和对k的余数。

#include <map>

#include <set>

#include <list>

#include <cmath>

#include <queue>

#include <stack>

#include <cstdio>

#include <vector>

#include <string>

#include <cctype>

#include <complex>

#include <cassert>

#include <utility>

#include <cstring>

#include <cstdlib>

#include <iostream>

#include <algorithm>

using namespace std;

typedef pair<int,int> PII;

typedef long long ll;

#define lson l,m,rt<<1

#define pi acos(-1.0)

#define rson m+1,r,rt<<11

#define All 1,N,1

#define read freopen("in.txt", "r", stdin)

const ll  INFll = 0x3f3f3f3f3f3f3f3fLL;

const int INF= 0x7ffffff;

const int mod =  1000000007;

int a,b,k;

int dp[35][110][110],bit[35];

int dfs(int i,int nmod,int smod,int e){

    if(i==0)return (nmod==0)&&(smod==0);

    if(!e&&dp[i][nmod][smod]!=-1)return dp[i][nmod][smod];

    int l=e?bit[i]:9;

    int num=0;

    for(int v=0;v<=l;++v){

        int tnmod=(nmod*10+v)%k;

        int tsmod=(smod+v)%k;

        num+=dfs(i-1,tnmod,tsmod,e&&(v==l));

    }

    return e?num:dp[i][nmod][smod]=num;

}

int solve(int x){

    int len=0;

    memset(dp,-1,sizeof(dp));

    while(x){

        bit[++len]=x%10;

        x/=10;

    }

    return dfs(len,0,0,1);

}

int main()

{

    int t;

    scanf("%d",&t);

    while(t--){

     scanf("%d%d%d",&a,&b,&k);

        if(k>=100)

           printf("0\n");

        else{

            printf("%d\n",solve(b)-solve(a-1));

        }

    }

return 0;

}

 

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