HDU 3271 SNIBB[数位DP]

Description

  As we know, some numbers have interesting property. For example, any even number has the property that could be divided by 2. However, this is too simple. 
  One day our small HH finds some more interesting property of some numbers. He names it the “Special Numbers In Base B” (SNIBB). Small HH is very good at math, so he considers the numbers in Base B. In Base B, we could express any decimal numbers. Let’s define an expression which describe a number’s “SNIBB value”.(Note that all the “SNIBB value” is in Base 10) 
  
    Here N is a non-negative integer; B is the value of Base. 
  For example, the “SNIBB value” of “1023” in Base “2” is exactly:10 
(As we know (1111111111)2=(1023)(10)) 
  Now it is not so difficult to calculate the “SNIBB value” of the given N and B. 
But small HH thinks that must be tedious if we just calculate it. So small HH give us some challenge. He would like to tell you B, the “SNIBB value” of N , and he wants you to do two kinds of operation: 
1.  What is the number of numbers (whose “SNIBB value” is exactly M) in the range [A,B]; 
2.  What it the k-th number whose “SNIBB value” is exactly M in the range [A,B]; (note that the first one is 1-th but not 0-th) 

Here M is given.

 

题意：给出一个SNIBB的定义，即一个十进制数N在B进制的情况下，每一位上的数加起来的和。2种询问，第一种：区间[A,B]内所有SNIBB恰好等于M的数的个数。第二种：区间[A,B]内第K个SNIBB恰好等于M的数是什么。

解法：数位DP，dfs(pos，base，state，limit) ，base表示进制，state只要记录每一位的和就可以了。找第K大的过程可以用二分实现。具体看代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<iostream>
#include<stdlib.h>
#include<set>
#include<map>
#include<queue>
#include<vector>
#include<bitset>
#pragma comment(linker, "/STACK:1024000000,1024000000")
template <class T>
bool scanff(T &ret){ //Faster Input
    char c; int sgn; T bit=0.1;
    if(c=getchar(),c==EOF) return 0;
    while(c!='-'&&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');
    if(c==' '||c=='\n'){ ret*=sgn; return 1; }
    while(c=getchar(),c>='0'&&c<='9') ret+=(c-'0')*bit,bit/=10;
    ret*=sgn;
    return 1;
}
#define inf 1073741823
#define llinf 4611686018427387903LL
#define PI acos(-1.0)
#define lth (th<<1)
#define rth (th<<1|1)
#define rep(i,a,b) for(int i=int(a);i<=int(b);i++)
#define drep(i,a,b) for(int i=int(a);i>=int(b);i--)
#define gson(i,root) for(int i=ptx[root];~i;i=ed[i].next)
#define tdata int testnum;scanff(testnum);for(int cas=1;cas<=testnum;cas++)
#define mem(x,val) memset(x,val,sizeof(x))
#define mkp(a,b) make_pair(a,b)
#define findx(x) lower_bound(b+1,b+1+bn,x)-b
#define pb(x) push_back(x)
using namespace std;
typedef unsigned long long  ll;
typedef pair<int,int> pii;


int dp[65][33][333];
int num[66];
ll dfs(int base,int pos,int m,bool lim){
    if(pos==0){
        if(m==0)return 1;
        return 0;
    }
    if(!lim&&~dp[base][pos][m])return dp[base][pos][m];
    int sum=0;
    int d=base-1;
    if(lim)d=num[pos];
    rep(i,0,d){
        if(i<=m)sum+=dfs(base,pos-1,m-i,lim&&d==i);
    }
    if(!lim)dp[base][pos][m]=sum;
    return sum;
}
int base,l,r,m;
int solve(int x){
    if(x<0)return 0;
    int len=0;
    for(;x;x/=base)num[++len]=x%base;
    return dfs(base,len,m,1);
}




int main(){
    mem(dp,-1);
    int op;
    int cas=0;
    while(scanf("%d",&op)!=EOF){
        printf("Case %d:\n",++cas);
        scanff(l);scanff(r);scanff(base);scanff(m);
        if(l>r)swap(l,r);
        if(op==1)printf("%d\n",solve(r)-solve(l-1));
        else{
            int k;scanff(k);
            int sum1=solve(l-1);
            while(r-l>1){
                int m=(int)((ll(l)+ll(r))>>1);
                int sum2=solve(m);
                if(sum2-sum1>=k)r=m;
                else l=m;
            }
            if(solve(l)-sum1==k)printf("%d\n",l);
            else if(solve(r)-sum1==k)printf("%d\n",r);
            else printf("Could not find the Number!\n");


        }

    }
    return 0;
}