Manthan, Codefest 17 E. Salazar Slytherin's Locket（数位DP）（进制）

传送门

题意：给你b,l,r，问你l到r在b进制中有多少个数是每一个数字都是偶数个。

做法：- -  数位dp很明显啊。dp[b][sta][c] 表示在b进制下，sta表示0 1状态， 1是偶数，0是奇数状态。c表示1-10的个数。 qd表示不一直都是0 。

小技巧：之后只判是奇数还是偶数，所以中途用^处理。每次记得回溯。

//china no.1
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <vector>
#include <iostream>
#include <string>
#include <map>
#include <stack>
#include <cstring>
#include <queue>
#include <list>
#include <stdio.h>
#include <set>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <cctype>
#include <sstream>
#include <functional>
#include <stdlib.h>
#include <time.h>
#include <bitset>
using namespace std;

#define pi acos(-1)
#define s_1(x) scanf("%d",&x)
#define s_2(x,y) scanf("%d%d",&x,&y)
#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define s_4(x,y,z,X) scanf("%d%d%d%d",&x,&y,&z,&X)
#define S_1(x) scan_d(x)
#define S_2(x,y) scan_d(x),scan_d(y)
#define S_3(x,y,z) scan_d(x),scan_d(y),scan_d(z)
#define PI acos(-1)
#define endl '\n'
#define srand() srand(time(0));
#define me(x,y) memset(x,y,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);
#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i<n;i++)
#define fOR(n,x,i) for(int i=n;i>=x;i--)
#define fOr(n,x,i) for(int i=n;i>x;i--)
#define W while
#define sgn(x) ((x) < 0 ? -1 : (x) > 0)
#define bug printf("***********\n");
#define db double
#define ll long long
#define mp make_pair
#define pb push_back
typedef long long LL;
const int INF=0x3f3f3f3f;
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
const int dx[]={-1,0,1,0,1,-1,-1,1};
const int dy[]={0,1,0,-1,-1,1,-1,1};
const int maxn=4e3+10;
const int maxx=4e5+10;
const double EPS=1e-8;
const double eps=1e-8;
const int mod=1e9+7;
template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}
template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}
template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}
template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}
template <class T>
inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}
while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}

inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;
while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}
else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}
if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}
if(IsN) num=-num;return true;}

void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}
void print(LL a){ Out(a),puts("");}
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
//cerr << "run time is " << clock() << endl;

int q,b;
int a[70];

LL dp[12][70][2][2][2][2][2][2][2][2][2][2][2];

bool check(int *t)
{
FOR(0,9,i)
{
if(t[i])
return 0;
}
return 1;
}
LL dfs(int pos,int qd,int sta,bool limit,int *c)
{
if(pos==-1)
return sta==1;//结束条件
if(!limit&&qd==0&&dp[b][pos][sta][c[0]][c[1]][c[2]][c[3]][c[4]][c[5]][c[6]][c[7]][c[8]][c[9]]!=-1)
return dp[b][pos][sta][c[0]][c[1]][c[2]][c[3]][c[4]][c[5]][c[6]][c[7]][c[8]][c[9]];
int up=limit ? a[pos] : (b-1);
LL tmp=0;
for(int i=0;i<=up;i++)
{
if(i==0&&qd==1) tmp+=dfs(pos-1,1,sta,limit&&i==a[pos],c);
else
{
c[i]^=1;
int st;
if(check(c))
st=1;
else st=0;
tmp+=dfs(pos-1,0,st,limit&&i==a[pos],c);
c[i]^=1;
}
}
if(!limit&&qd==0) dp[b][pos][sta][c[0]][c[1]][c[2]][c[3]][c[4]][c[5]][c[6]][c[7]][c[8]][c[9]]=tmp;
return tmp;
}

LL solve(LL x)
{
int pos=0;
int t[12];
me(t,0);
while(x)
{
a[pos++]=x%b;
x/=b;
}
return dfs(pos-1,1,0,true,t);
}
int main()
{
//freopen( "in.txt" , "r" , stdin );
S_1(q);
me(dp,-1);
W(q--)
{
LL l,r;
S_3(b,l,r);
print(solve(r)-solve(l-1));
}
}