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-;
 const double pi = acos(-1.0);
 const double inf = ~0u>>;
 int dp[][][][];
 int d[];
 int dfs(int i,int e,int sum,int m,int r)
 {
     if(i==-)
     {
         return r==&&(sum==m);
     }
     if(!e&&dp[i][sum][m][r]!=-) return dp[i][sum][m][r];
     int mk = e?d[i]:;
     int ans = ;
     for(int j =  ; j <= mk ; j++)
     {
         ans+=dfs(i-,e&&j==mk,sum+j,m,(r*+j)%m);
     }
     return e?ans:dp[i][sum][m][r] = ans;
 }
 int cal(int x)
 {
     int g = ;
     if(x<=) return ;
     while(x)
     {
         d[g++] = x%;
         x/=;
     }
     int ans = ;
     for(int i =  ; i <=  ; i++)
     {

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