hdu4389 X mod f(x)

链接

这个题因为总和加起来是比较小的9*9 = 81  这样可以保留前面枚举的数对所有的可能出现的和的余数，然后依次向下找。

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stdlib.h>
#include<vector>
#include<cmath>
#include<queue>
#include<set>
using namespace std;
#define N 100000
#define LL long long
#define INF 0xfffffff
const double eps = 1e-8;
const double pi = acos(-1.0);
const double inf = ~0u>>2;
int dp[10][82][82][82];
int d[10];
int dfs(int i,int e,int sum,int m,int r)
{
    if(i==-1)
    {
        return r==0&&(sum==m);
    }
    if(!e&&dp[i][sum][m][r]!=-1) return dp[i][sum][m][r];
    int mk = e?d[i]:9;
    int ans = 0;
    for(int j = 0 ; j <= mk ; j++)
    {
        ans+=dfs(i-1,e&&j==mk,sum+j,m,(r*10+j)%m);
    }
    return e?ans:dp[i][sum][m][r] = ans;
}
int cal(int x)
{
    int g = 0;
    if(x<=0) return 0;
    while(x)
    {
        d[g++] = x%10;
        x/=10;
    }
    int ans = 0;
    for(int i = 1 ; i <= 81 ; i++)
    {

        int k = dfs(g-1,1,0,i,0);
        ans+=k;
    }
    return ans;
}
int main()
{
    int t,a,b;
    cin>>t;
    int kk = 0;
    memset(dp,-1,sizeof(dp));
    while(t--)
    {
        cin>>a>>b;
        printf("Case %d: ",++kk);
        cout<<cal(b)-cal(a-1)<<endl;
    }
    return 0;
}