hdu 3271 SNIBB 数位DP+二分

思路：dp[i][j]:表示第i位在B进制下数字和。

用二分找第k个数！

代码如下：

 

 1 #include<iostream>

 2 #include<stdio.h>

 3 #include<algorithm>

 4 #include<iomanip>

 5 #include<cmath>

 6 #include<cstring>

 7 #include<vector>

 8 #define ll __int64

 9 using namespace std;

10 int dp[45][310],b,m,bit[45];

11 int dfs(int pos,int mm,bool f)

12 {

13     if(pos==-1) return mm==m;

14     if(!f&&dp[pos][mm]!=-1) return dp[pos][mm];

15     int ans=0;

16     int e=f?bit[pos]:(b-1);

17     for(int i=0;i<=e;i++){

18         ans+=dfs(pos-1,mm+i,f&&i==bit[pos]);

19     }

20     if(!f) dp[pos][mm]=ans;

21     return ans;

22 }

23 int solve(int n)

24 {

25     if(n<=0) return n==m;

26     int s=0;

27     while(n){

28         bit[s++]=n%b;

29         n/=b;

30     }

31     return dfs(s-1,0,1);

32 }

33 int main(){

34     int q,x,y,n,k,ca=0;

35     while(scanf("%d%d%d%d%d",&q,&x,&y,&b,&m)!=EOF){

36         memset(dp,-1,sizeof(dp));

37         printf("Case %d:\n",++ca);

38         if(x>y) swap(x,y);

39         if(q==1) printf("%d\n",solve(y)-solve(x-1));

40         else{

41             scanf("%d",&k);

42             int low=solve(x-1);

43             int high=solve(y);

44             if(high-low<k){

45                 puts("Could not find the Number!");

46                 continue;

47             }

48             int l=x,r=y,mid,ans=0;

49             while(l<=r){

50                 mid=(int)(((ll)l+(ll)r)>>1);

51                 int t=solve(mid);

52                 if(t-low>=k) r=mid-1,ans=mid;

53                 else l=mid+1;

54             }

55             printf("%d\n",ans);

56         }

57     }

58     return 0;

59 }

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