URAL 1057 数位DP

Create a code to determine the amount of integers, lying in the set [ X; Y] and being a sum of exactly K different integer degrees ofB.

Example. Let X=15, Y=20, K=2, B=2. By this example 3 numbers are the sum of exactly two integer degrees of number 2:

17 = 2 ⁴+2 ⁰, 
18 = 2 ⁴+2 ¹, 
20 = 2 ⁴+2 ².

题意:  区间[X,Y]中。  问能构成 取 K 个不同的 B的几次方的数有多少个。 

也就是区间[1,Y] - [1,X-1];

数位DP题吧。。每位判断的的时候 最多取一个1.。

其中state是判断是否跟原先数相等。。

#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>using namespace std;int n,m,k,b;int dp[44][44]; // dp[k][len]  , 表示取几个数 跟 表示取到第几位, int vis[44][44];int num[44],kn;long long dfs(int k , int len, bool state){if(k == 0 )return (len >= -1 );if(k == 0 || len < -1) return 0;if(k > 0 && len < 0) return 0;if(!state && dp[k][len] != -1)return dp[k][len];long long ans = 0;int end = state ? num[len] : 1;//printf("end=%d\n",end);if(end > 1) end = 1,state = false;  // 最多去一个 B^len次 for(int i = 0;i <= end; i++){ans += dfs( k - i, len - 1 , state && i == end);}if(!state)dp[k][len] = ans;//printf(" dfs(%d , %d , %d ) = %lld\n",k , len, (int)state , ans);return ans;}int solve(int n){memset(dp,-1,sizeof(dp));int tn = n; kn = 0;while(tn > 0){num[kn++] = tn % b;tn /= b;}//for(int i = 0; i < kn;i++){//printf("num[%d]=%d\n",i,num[i]);//}long long ans = dfs(k, kn-1, num[kn-1] == 1);return ans;}int main(){while(scanf("%d %d",&n,&m)!=EOF){scanf("%d %d",&k,&b);printf("%d\n",solve(m)-solve(n-1));}return 0;}