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hdu 4389 数位DP 模板化代码 结合记忆化搜索优化

2016-02-16 23:14 567 查看


数位DP

用的是之前的模板。
之前写过的汇总:/article/8874680.html
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);
}
}
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