topcoder: TCHS10 Championship Round DIV 1 (lucky number third level)

Problem Statement

 

The digits 4 and 7 are lucky digits, and all other digits are unlucky. A third level lucky number is a positive integer that is divisible by at least one lucky digit. John and Brus have written down all the third level lucky numbers betweena and b, inclusive, and now they would like to count the number of occurrences of each digit between 0 and 9, inclusive, in the numbers they have written.

Return a vector<long long> containing exactly 10 elements, where the i-th element (0-based) is the total number of occurrences of digit i.

Definition

 Class:TheLuckyNumbersLevelThreeMethod:findParameters:long long, long longReturns:vector<long long>Method signature:vector<long long> find(long long a, long long b)(be sure your method is public)

Limits

 Time limit (s):2.000Memory limit (MB):64

Constraints

-a will be between 1 and 10^16, inclusive.-b will be between a and 10^16, inclusive.

数位dp的题居然也能压轴～～服了。

题意：列出区间[a,b]所有能被4或者7整除的数，统计这些数含有多少个digit s，s:0,1,...,9

example:

a = 1, b = 10

Returns: {0, 0, 0, 0, 1, 0, 0, 1, 1, 0 }

There are three third level lucky numbers in this range - 4, 7, 8.

tricky: 传统的数位dp一般数统计满足条件的数的个数，但是这个题统计的是满足条件的数中digit 0-9的个数。

一开始，如果不仔细分析到这里就很容易跳进死胡同，其实此题可以很简单的转换到常规数位dp。

换一种问法：区间[a,b]中所有满足条件的数x：

x能被4或者7整除的

x含有k个1

这里就可以枚举 k 和digit 0-9。

然后，设计数位dp的递归函数就比较方便了。

#include<cmath>#include<cstdlib>#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<vector>using namespace std;#define FOR(i,a,b) for(int i=a;i<b;i++)#define FORE(i,a,b) for(int i=a;i<=b;i++)#define MST(a,b) memset(a,b,sizeof(a))class TheLuckyNumbersLevelThree{public:int cur,n;int d[17];int tmpsum;long long dp[17][17][7][4];int mod4[17],mod7[17];TheLuckyNumbersLevelThree(){mod4[0] = mod7[0] = 1;FOR(i,1,17) mod4[i] = mod4[i-1]*10%4;FOR(i,1,17) mod7[i] = mod7[i-1]*10%7;tmpsum = 0;}long long dfs(int i,int c,int u,int v,bool flag) {if(i==-1){if(!u || !v)return c;else return 0;}if(!flag && tmpsum && dp[i][c][u][v]!=-1 )return  dp[i][c][u][v];long long ans = 0;int lim = (flag?d[i]:9);FORE(s,0,lim){tmpsum += s;ans += dfs(i-1,c+((s==cur) && tmpsum),(u+s*mod7[i])%7,(v+s*mod4[i])%4,flag && (s==lim));tmpsum -= s;}if(!flag && tmpsum) dp[i][c][u][v] = ans;return ans;}vector<long long> find(long long a, long long b){long long sa,sb,x;vector<long long>  ans;FOR(i,0,10){cur = i;for(x = a-1,n = 0;x;x/=10)d[n++]=x%10;MST(dp,-1);sa = dfs(n-1,0,0,0,1);for(x = b,n = 0;x;x/=10)d[n++]=x%10;MST(dp,-1);sb = dfs(n-1,0,0,0,1);ans.push_back(sb-sa);}return ans;}};