hdu 4389 数位DP 模板化代码 结合记忆化搜索优化

数位DP

用的是之前的模板。

之前写过的汇总： http://blog.youkuaiyun.com/techmonster/article/details/50284375 

模板： (强调一点，我的数位DP模板是结合记忆话搜索优化的，博客中之前写过的数位DP题，都是用这个模板)

int dfs(int top, int s, bool e) 
{
    if (top==0) return s==target_s;
    if (!e && dp[top][s]!=-1) return dp[top][s];
    int i,res = 0;
    int u = e?num[top]:9;
    for (i = 0; i <= u; ++i)
        res += dfs(top-1, new_s(s, i), e&&i==u);
    return e?res:dp[top][s]=res;
}

dp[i][j][k][l]   i表示第几位数，j表示各位数总和，k表示余数，l 表示除多少
题目代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
#include<queue>
#include<cctype>
#define MS(x,y) memset(x,y,sizeof(x))
using namespace std;
const int MAXN = 130;
//const int N = 5010;
const double eps = 1e-10;
void fre(){freopen("t.txt","r",stdin);}
typedef long long LL;
typedef unsigned long long ULL;


int len,dp[10][82][82][82],num[15];
void getnum(int x)
{
    len = 0;
    while(x)
    {
        num[++len] = x%10;
        x/=10;
    }
}
int dfs(int top, int sum, int mo, int  div,bool e)
{
    if (top==0) return mo==0 && sum == div;
    if (!e && dp[top][sum][mo][div]!=-1) return dp[top][sum][mo][div];
    int i,res = 0,nsum,ndiv,nmo;
    int u = e?num[top]:9;
    for (i = 0; i <= u; ++i)
    {
        nsum = sum+i;
        res += dfs(top-1, nsum, (mo*10+i)%div, div, e&&i==u);
    }
    return e?res:dp[top][sum][mo][div]=res;
}
int main()
{
   // fre();
    int T,a,b,i,t = 0;
    scanf("%d",&T);MS(dp,-1);
    while(T--)
    {
        scanf("%d%d",&a,&b);
        int tem = 0;
        getnum(b);
        for(i = 1; i < 82; ++i) tem +=dfs(len,0,0,i,1);
        getnum(a-1);
        for(i = 1; i < 82; ++i) tem -=dfs(len,0,0,i,1);
        printf("Case %d: %d\n",++t,tem);
    }
}