hdu6156 Palindrome Function 数位dp

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题目：求L~R所有的数的l~r进制的f(n，k), 如果n在k进制下是回文串那么f(n，k) = k，否则等于1。

思路：数位dp

代码：

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<algorithm>#include<ctime>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<list>#include<numeric>using namespace std;#define LL long long#define ULL unsigned long long#define INF 0x3f3f3f3f3f3f3f3f#define mm(a,b) memset(a,b,sizeof(a))#define PP puts("*********************");template<class T> T f_abs(T a){ return a > 0 ? a : -a; }template<class T> T gcd(T a, T b){ return b ? gcd(b, a%b) : a; }template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}// 0x3f3f3f3f3f3f3f3fLL dp[40][105][105][2];int dight[105],num[105];LL dfs(int base,int pos,int len,int sta,bool limit){    if(pos<0){        if(sta) return base;        else return 1;    }    if(dp[base][pos][len][sta]!=-1&&!limit)        return dp[base][pos][len][sta];    int up=limit?dight[pos]:(base-1);    LL sum=0;    for(int i=0;i<=up;i++){        num[pos]=i;        if(i==0&&pos==len-1){            sum+=dfs(base,pos-1,len-1,sta,limit&&(i==up));        }        else if(pos<len/2){            if(i==num[len-pos-1])                sum+=dfs(base,pos-1,len,sta,limit&&(i==up));            else                sum+=dfs(base,pos-1,len,0,limit&&(i==up));        }        else            sum+=dfs(base,pos-1,len,sta,limit&&(i==up));    }    if(!limit) dp[base][pos][len][sta]=sum;    return sum;}LL solve(int n,int base){    if(n==0)        return base;    int len=0;    do{        dight[len++]=n%base;        n/=base;    }while(n>0);    return dfs(base,len-1,len,1,1);}int main(){    int T,cas=0,L,R,l,r;    mm(dp,-1);    scanf("%d",&T);    while(T--){        scanf("%d%d%d%d",&L,&R,&l,&r);        LL ans=0;        for(int i=l;i<=r;i++)            ans+=solve(R,i)-solve(L-1,i);        printf("Case #%d: %lld\n",++cas,ans);    }    return 0;}

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